limx→0x2cosx1−cosx is (a) 2 (b) 3 2 (c) − 3 2 (d) 1

This is an objective Type Questions as classified in NCERT Exemplar Sol: Given that limx→0x2cosx1−cosx =limx→0x2cosx2sin2x2=−1 [?1−cosx=2sin2x2] =limx→0x24×4cosx2sin2x2=limx→0x2→0(x2)2×2cosxsin2x2 =limx2→0(x2sinx2)2×2cosx=2cos0=2×1=2 [?limx→0xsinx=1]Hence, the correct option is (a).

Kindly consider the following limx→1xm−1xn−1 is (a) 1 (b) m n (c) m n − 1 (d) m 2 n 2

This is an objective Type Questions as classified in NCERT Exemplar Sol: Given that limx→1xm−1xn−1 =limx→1xm−(1)mx−1xn−(1)nx−1=m(1)m−1n(1)n−1=mn [?limx→axn−anx−a=n.an−1]Hence, the correct option is (b).

Kindly consider the following limθ→01−cos4θ1−cos6θ is (a) 4 9 (b) 1 2 (c) − 1 2 (d) -1

This is an objective Type Questions as classified in NCERT Exemplar Sol: Given that limθ→01−cos4θ1−cos6θ =limθ→02sin22θ2sin23θ [?1−cosθ=2sin2θ2] =limθ→0sin22θsin23θ=limθ→0[sin2θsin3θ]2=limθ→02θ→03θ→0[sin2θ2θ×2θsin3θ3θ×3θ]2 =[2θ3θ]2=(23)2=49Hence, the correct option is (a).

Kindly consider the following limx→0cosec x−cot xx is (a) − 1 2 (b) 1 (c) 1 2 (d) 1

This is an objective Type Questions as classified in NCERT Exemplar Sol: Given that limx→0cosec x−cot xx =limx→01sinx−cos xsinxx=limx→01−cosxxsinx=limx→02sin2x2x.2sinx2cosx2 =limx→0sinx2xcosx2=limx→0tanx2x=limx→0tanx22×x2=12×1=12 [?limx→0tanxx=1]Hence, the correct option is (c).

Kindly consider the following ax+bcx+d

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Let f(x)=ax+bcx+d …(i)⇒ f(x+Δx)=a(x+Δx)+bc(x+Δx)+d …(ii)Subtracting eqn.(i) from eqn.(ii) f(x+Δx)−f(x)=a(x+Δx)+bc(x+Δx)+d−ax+bcx+dDividing both sides by Δx and take the we get limΔx→0f(x+Δx)−f(x)Δx=limΔx→0a(x+Δx)+bc(x+Δx)+d−ax+bcx+dΔx f'(x)=limΔx→0(ax+aΔx+b)(cx+d)−(ax+b)(cx+cΔx+d)[c(x+Δx)+d](cx+d).Δx =limΔx→0acx2+acΔx.x+bcx+adx+adΔx+bd−acx2−acΔx.x−adx−bcx−bc.Δx−bd(cx+cΔx+d)(cx+d)Δx =limΔx→0(ad−bc)Δx(cx+cΔx+d)(cx+d).Δx=limΔx→0(ad−bc)(cx+c.Δx+d)(cx+d)Taking limit , we have =(ad−bc)(cx+d)(cx+d)=(ad−bc)(cx+d)2

Kindly consider the following x23

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Let f(x)=x2/3 …(i)⇒ f(x+Δx)=(x+Δx)2/3 …(ii)Subtracting eqn.(i) from eqn.(ii) f(x+Δx)−f(x)=(x+Δx)2/3−x2/3Dividing both sides by Δx and take the we get limΔx→0f(x+Δx)−f(x)Δx=limΔx→0(x+Δx)2/3−x2/3Δx f'(x)=limΔx→0x2/3[1+Δxx]2/3−x2/3Δx =limΔx→0x2/3[(1+Δxx)2/3−1]Δx =limΔx→0x2/3[(1+23.Δxx+…)−1]Δx [Expanding by Binomial theorem and rejectingthe higher powers of Δx as Δx→0] =limΔx→0x2/3.23.ΔxxΔx=23x2/3−1=23x−1/3.

Kindly consider the following l i m x → 3 x 2 − 9 x − 3

This is Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that limx→0x2−9x−3=limx→0 (x+3) (x−3) (x−3)=limx→0x+3Taking , we have 3+3=6.

Kindly consider the following limx→1/24x2−12x−1

This is Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that limx→124x2−12x−1=limx→12(2x)2−(1)22x−1=limx→12(2x−1)(2x+1)2x−1 =limx→122x+1Taking , we have 2×12+1=1+1=2

Kindly consider the following limh→0x+h−xh

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that limh→0x+h−xh =limh→0x+h−xh×x+h+xx+h+x [Rationalizing the denominator ] =limh→0x+h−xh[x+h+x]=limh→0hh[x+h+x] =limh→01x+h+xTaking limit , we have 1x+x=12x.

Kindly consider the following limx→0(x+2)13−213x

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that limx→0(x+2)13−213xPut x+2=y ⇒x=y−2 =limy−2→0y13−213y−2=limy→2y13−213y−2 =13.(2)13−1=13.2−23 [ Usinglimx→axn−anx−a=n.an−1]

Kindly consider the following limx→1(1+x)6−1(1+x)2−1

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that limx→0(1+x)6−1(1+x)2−1Dividing the numerator and denominator by x, we get =limx→0(1+x)6−1x(1+x)2−1xPut 1+x=y ⇒x=y−1 =limy−1→0y6−(1)6y−1y2−(1)2y−1=limy→1y6−(1)6y−1limy→1y2−(1)2y−1 [limx→af(x)g(x)=limx→af(x)limx→ag(x)] =6.(1)6−12.(1)2−1=62=3 [ limx→axn−anx−a=n.an−1]

If f(x)=tanxx−π , then limx→πf(x)=_________ .

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: Given f (x)=limx→π−tan (π−x)x−π =limπ−x→0−tan (π−x)− (π−x)=1Hence, the value of the filler is 1.

limx→0sin(mx)cot x/ √ 3 = 2 then m = __________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: Given that limx→0(sinmxcotx3)=2⇒ limx→0∴mx→0sinmxmx×mx limx→0(cotx3)=2⇒ 1×mx limx→01tanx3=2⇒ limx→0 mx×x3x3.tanx3=2⇒ mxx3×(1)=2 ⇒3m=2 ⇒m=23=233.Hence, the value of the filler is 233.

Kindly consider the following If y=1+x1!+x22!+x33!+... , then dydx=_________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: Given that y=1+x1!+x22!+x33!+… dydx=0+11!+2x2!+3x23!+… =1+x1!+x22!+x33!+…=yHence, the value of the filler is y.

Kindly consider the following limx→3x[x]=_________ .

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: G i v e n l i m x → 3 + x [ x ] = l i m h → 0 ( 3 + h ) [ 3 + h ] = 1 H e n c e , t h e v a l u e o f t h e f i l l e r i s 1 .

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen: Overview, Questions, Preparation

Updated on Jul 7, 2025 16:04 IST

By alok kumar singh, Executive Content Operations

Table of contents

Limits and Derivatives Short Answer Type Questions
Limits and Derivatives Long Answer Type Questions
Limits and Derivatives Objective Type Questions
Limits and Derivatives Fill in the blanks

Limits and Derivatives Short Answer Type Questions

l i m_{x \to 3} \frac{x^{2} - 9}{x - 3}

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{x^{2} - 9}{x - 3} = \underset{x \to 0}{l i m} \frac{(x + 3) (x - 3)}{(x - 3)} = \underset{x \to 0}{l i m} x + 3 \\ T a k i n g, w e h a v e 3 + 3 = 6 . \end{array}$

2. $l i m_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1}$

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{1}{2}}{l i m} \frac{4 x^{2} - 1}{2 x - 1} = \underset{x \to \frac{1}{2}}{l i m} \frac{{(2 x)}^{2} - {(1)}^{2}}{2 x - 1} = \underset{x \to \frac{1}{2}}{l i m} \frac{(2 x - 1) (2 x + 1)}{2 x - 1} \\ = \underset{x \to \frac{1}{2}}{l i m} 2 x + 1 \\ T a k i n g, w e h a v e \\ 2 \times \frac{1}{2} + 1 = 1 + 1 = 2 \end{array}$

Commonly asked questions

Kindly consider the following

$l i m_{x \to 3} \frac{x^{2} - 9}{x - 3}$

This is Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$l i m_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1}$

This is Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$l i m_{h \to 0} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \times \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e denominator] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g limit, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x} \\ P u t x + 2 = y \Rightarrow x = y - 2 \\ = \underset{y - 2 \to 0}{l i m} \frac{y^{\frac{1}{3}} - 2^{\frac{1}{3}}}{y - 2} = \underset{y \to 2}{l i m} \frac{y^{\frac{1}{3}} - 2^{\frac{1}{3}}}{y - 2} \\ = \frac{1}{3} . {(2)}^{\frac{1}{3} - 1} = \frac{1}{3} . 2^{\frac{- 2}{3}} [Using \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \end{array}$

Kindly consider the following

$l i m_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y x, w e g e t \\ = \underset{x \to 0}{l i m} \frac{\frac{{(1 + x)}^{6} - 1}{x}}{\frac{{(1 + x)}^{2} - 1}{x}} \\ P u t 1 + x = y \Rightarrow x = y - 1 \\ = \underset{y - 1 \to 0}{l i m} \frac{\frac{y^{6} - {(1)}^{6}}{y - 1}}{\frac{y^{2} - {(1)}^{2}}{y - 1}} = \frac{\underset{y \to 1}{l i m} \frac{y^{6} - {(1)}^{6}}{y - 1}}{\underset{y \to 1}{l i m} \frac{y^{2} - {(1)}^{2}}{y - 1}} [\underset{x \to a}{l i m} \frac{f (x)}{g (x)} = \frac{\underset{x \to a}{l i m} f (x)}{\underset{x \to a}{l i m} g (x)}] \\ = \frac{6 . {(1)}^{6 - 1}}{2 . {(1)}^{2 - 1}} = \frac{6}{2} = 3 [\underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \end{array}$

Kindly consider the following

$\underset{x \to a}{l i m} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(x - a)}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to a}{l i m} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(x - a)} \\ = \underset{2 + x \to a + 2}{l i m} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(2 + x) - (a + 2)} \\ = \frac{5}{2} {(a + 2)}^{\frac{5}{2} - 1} = \frac{5}{2} {(a + 2)}^{\frac{3}{2}} [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \end{array}$

Kindly consider the following

$l i m_{x \to 1} \frac{x^{4} - \sqrt[]{x}}{\sqrt[]{x} - 1}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{4} - \sqrt[]{x}}{\sqrt[]{x} - 1} \\ = \underset{x \to 1}{l i m} \frac{\sqrt[]{x} [{(x)}^{7 / 2} - 1]}{\sqrt[]{x} - 1} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y x - 1 \\ = \underset{x \to 1}{l i m} \frac{\frac{\sqrt[]{x} [{(x)}^{7 / 2} - {(1)}^{7 / 2}]}{x - 1}}{\frac{{(x)}^{1 / 2} - {(1)}^{1 / 2}}{x - 1}} \\ = \underset{x \to 1}{l i m} \frac{\frac{{(x)}^{7 / 2} - {(1)}^{7 / 2}}{x - 1}}{\frac{{(x)}^{1 / 2} - {(1)}^{1 / 2}}{x - 1}} \times \underset{x \to 1}{l i m} \sqrt[]{x} [? \underset{x \to a}{l i m} f (x) . g (x) = \underset{x \to a}{l i m} f (x) . \underset{x \to a}{l i m} g (x)] \\ = \frac{\frac{7}{2} {(1)}^{7 / 2 - 1}}{\frac{1}{2} {(1)}^{1 / 2 - 1}} \times \sqrt[]{1} = \frac{7 / 2}{1 / 2} = 7 . \end{array}$

Kindly consider the following

$l i m_{x \to 2} \frac{x^{2} - 4}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 2}{l i m} \frac{x^{2} - 4}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}} \\ R a t i o n a l i z i n g t h e denominator w e g e t \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2)}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}} \times \frac{\sqrt[]{3 x - 2} + \sqrt[]{x + 2}}{\sqrt[]{3 x - 2} + \sqrt[]{x + 2}} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{3 x - 2 - x - 2} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2 x - 4} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2 (x - 2)} = \underset{x \to 2}{l i m} \frac{(x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2} \\ T a k i n g limits, w e h a v e \\ = \frac{(2 + 2) (\sqrt[]{6 - 2} + \sqrt[]{2 + 2})}{2} = \frac{4 (2 + 2)}{2} = 8 \end{array}$

Kindly consider the following

$l i m_{x \to \sqrt[]{2}} \frac{x^{4} - 4}{x^{2} + 3 \sqrt[]{2} x - 8}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \sqrt[]{2}}{l i m} \frac{x^{4} - 4}{x^{2} + 3 \sqrt[]{2} x - 8} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x^{2} - 2) (x^{2} + 2)}{x^{2} + 4 \sqrt[]{2} x - \sqrt[]{2} x - 8} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x - \sqrt[]{2}) (x^{2} + 2)}{x (x + 4 \sqrt[]{2}) - \sqrt[]{2} (x + 4 \sqrt[]{2})} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x - \sqrt[]{2}) (x^{2} + 2)}{(x + 4 \sqrt[]{2}) (x - \sqrt[]{2})} = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x^{2} + 2)}{(x + 4 \sqrt[]{2})} \\ T a k i n g, w e h a v e \\ = \frac{(\sqrt[]{2} + \sqrt[]{2}) (2 + 2)}{\sqrt[]{2} + 4 \sqrt[]{2}} = \frac{2 \sqrt[]{2} \times 4}{5 \sqrt[]{2}} = \frac{8}{5} . \end{array}$

Kindly consider the following

$l i m_{x \to 3} \frac{x^{7} - 2 x^{5} - 1}{x^{3} + 3 x^{2} + 2}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{7} - 2 x^{5} - 1}{x^{3} + 3 x^{2} + 2} [\frac{0}{0} f o r m] \\ = \underset{x \to 1}{l i m} \frac{x^{7} - x^{5} - x^{5} - 1}{x^{3} - x^{2} - 2 x^{2} + 2} = \underset{x \to 1}{l i m} \frac{x^{5} (x^{2} - 1) - 1 (x^{5} - 1)}{x^{2} (x - 1) - 2 (x^{2} - 1)} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y (x - 1) w e g e t \\ = \underset{x \to 1}{l i m} \frac{x^{5} (\frac{x^{2} - 1}{x - 1}) - 1 (\frac{x^{5} - 1}{x - 1})}{x^{2} (\frac{x - 1}{x - 1}) - 2 (\frac{x^{2} - 1}{x - 1})} = \frac{\underset{x \to 1}{l i m} x^{5} (x + 1) - \underset{x \to 1}{l i m} (\frac{x^{5} - {(1)}^{5}}{x - 1})}{\underset{x \to 1}{l i m} x^{2} - 1 - 2 \underset{x \to 1}{l i m} (x + 1)} \\ = \frac{1 (2) - 5 . {(1)}^{5 - 1}}{1 - 2 (2)} = \frac{2 - 5}{1 - 4} = \frac{- 3}{- 3} = 1 \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}} \\ R a t i o n a l i z i n g t h e denominator, w e g e t \\ = \underset{x \to 0}{l i m} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}} \times \frac{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}}{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}} \\ = \underset{x \to 0}{l i m} \frac{(1 + x^{3}) - (1 - x^{3})}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} = \underset{x \to 0}{l i m} \frac{1 + x^{3} - 1 + x^{3}}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} \\ = \underset{x \to 0}{l i m} \frac{2 x^{3}}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} = \underset{x \to 0}{l i m} \frac{2 x}{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}} = 0 \end{array}$

Kindly consider the following

$l i m_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to - 3}{l i m} \frac{x^{3} + 27}{x^{5} + 243} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator, b y x - 3 w e g e t \\ = \underset{x \to - 3}{l i m} \frac{\frac{x^{3} + {(3)}^{3}}{x - 3}}{\frac{x^{5} + {(3)}^{5}}{x - 3}} = \frac{\underset{x \to - 3}{l i m} (\frac{x^{3} - {(- 3)}^{3}}{x - 3})}{\underset{x \to - 3}{l i m} (\frac{x^{5} - {(- 3)}^{5}}{x - 3})} [? \underset{x \to a}{l i m} \frac{f (x)}{g (x)} = \frac{\underset{x \to a}{l i m} f (x)}{\underset{x \to a}{l i m} g (x)}] \\ = \frac{3 {(- 3)}^{3 - 1}}{5 {(- 3)}^{5 - 1}} [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ = \frac{3 \times {(- 3)}^{2}}{5 \times {(- 3)}^{4}} = \frac{1}{5 \times 3} = \frac{1}{15} \end{array}$

Kindly consider the following

$l i m_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{1}{2}}{l i m} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}) \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{(8 x - 3) (2 x + 1) - (4 x^{2} + 1)}{(4 x^{2} - 1)}] \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{16 x^{2} - 6 x + 8 x - 3 - 4 x^{2} - 1}{4 x^{2} - 1}] \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{12 x^{2} + 2 x - 4}{4 x^{2} - 1}] = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (6 x^{2} + x - 2)}{4 x^{2} - 1} \\ = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (6 x^{2} + 4 x - 3 x - 2)}{(2 x + 1) (2 x - 1)} = \underset{x \to \frac{1}{2}}{l i m} \frac{2 [2 x (3 x + 2) - 1 (3 x + 2)]}{(2 x + 1) (2 x - 1)} \\ = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (3 x + 2) (2 x - 1)}{(2 x + 1) (2 x - 1)} = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (3 x + 2)}{(2 x + 1)} \\ T a k i n g limit, w e h a v e \\ = \frac{2 (3 \times \frac{1}{2} + 2)}{2 \times \frac{1}{2} + 1} = \frac{2 (\frac{7}{2})}{2} = \frac{7}{2} \end{array}$

Find $n$ , if $l i m_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80$ , $n \inN$
.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 2}{l i m} \frac{x^{n} - 2^{n}}{x - 2} = 80 \\ n . {(2)}^{n - 1} = 80 [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ \Rightarrow n \times 2^{n - 1} = 5 \times {(2)}^{5 - 1} \\ ∴ n = 5 \end{array}$

Kindly consider the following

$l i m_{x \to a} \frac{s i n 3 x}{s i n 7 x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 3 x}{s i n 7 x} \\ = \underset{x \to 0}{l i m} \frac{\frac{s i n 3 x}{3 x} \times 3 x}{\frac{s i n 7 x}{7 x} \times 7 x} = \frac{\underset{3 x \to 0}{l i m} (\frac{s i n 3 x}{3 x})}{\underset{3 x \to 0}{l i m} (\frac{s i n 7 x}{7 x})} \times \frac{3}{7} \\ = \frac{1}{1} \times \frac{3}{7} = \frac{3}{7} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{s i n^{2} 2 x}{s i n^{2} 4 x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 4 x} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 2 (2 x)} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{4 s i n^{2} 2 x . c o s^{2} 2 x} [s i n 2 x = 2 s i n x c o s x] \\ = \underset{x \to 0}{l i m} \frac{1}{4 c o s^{2} 2 x} \\ T a k i n g limit, w e h a v e \\ = \frac{1}{4 . c o s^{2} 0} = \frac{1}{4} \end{array}$

Kindly consider the following
$l i m_{x \to 0} \frac{1 - c o s 2 x}{x^{2}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{1 - c o s 2 x}{x^{2}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n^{2} x}{x^{2}} [c o s 2 x = 1 - 2 s i n^{2} x] \\ = \underset{x \to 0}{l i m} 2 {(\frac{s i n x}{x})}^{2} = 2 \times 1 = 2 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{2 s i n x - s i n 2 x}{x^{3}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{2 s i n x - s i n 2 x}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x - 2 s i n x c o s x}{x^{3}} = \underset{x \to 0}{l i m} \frac{2 s i n x (1 - c o s x)}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x}{x} (\frac{1 - c o s x}{x^{2}}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (\frac{2 s i n^{2} x / 2}{x^{2}}) \\ = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (2 \frac{s i n^{2} x / 2}{\frac{x^{2}}{4}} \times \frac{1}{4}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) 2 {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} \times \frac{1}{4} \\ = \underset{x \to 0}{l i m} \frac{4}{4} (\frac{s i n x}{x}) \underset{\frac{x}{2} \to 0}{l i m} {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} = 1.1 . {(1)}^{2} = 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{1 - c o s m x}{1 - c o s n x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{1 - c o s m x}{1 - c o s n x} \\ = \underset{x \to 0}{l i m} (\frac{1 - 1 + 2 s i n^{2} \frac{m x}{2}}{1 - 1 + 2 s i n^{2} \frac{n x}{2}}) [? c o s m x = 1 - 2 s i n^{2} \frac{m x}{2}] \\ = \underset{x \to 0}{l i m} (\frac{2 s i n^{2} \frac{m x}{2}}{2 s i n^{2} \frac{n x}{2}}) = \underset{x \to 0}{l i m} {(\frac{s i n \frac{m x}{2}}{s i n \frac{n x}{2}})}^{2} = \frac{\underset{x \to 0}{l i m} (\frac{s i n \frac{m x}{2}}{\frac{m x}{2}} \times \frac{m x}{2})}{\underset{x \to 0}{l i m} (\frac{s i n \frac{n x}{2}}{\frac{n x}{2}} \times \frac{n x}{2})} \\ = \frac{1 . \frac{m^{2} x^{2}}{4}}{1 . \frac{n^{2} x^{2}}{4}} = \frac{m^{2}}{n^{2}} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following
$\underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)} \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2 s i n^{2} 3 x}}{\sqrt[]{2} (\frac{π}{3} - x)} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2} s i n 3 x}{\sqrt[]{2} (\frac{π}{3} - x)} = \underset{π - 3 x \to 0}{l i m} \frac{3 . s i n (π - 3 x)}{π - 3 x} \\ = 3 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following
$l i m_{x \to \frac{π}{4}} \frac{s i n x - c o s x}{x - \frac{π}{4}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{s i n x - c o s x}{x - \frac{π}{4}} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (\frac{1}{\sqrt[]{2}} s i n x - \frac{1}{\sqrt[]{2}} c o s x)}{x - \frac{π}{4}} = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (c o s \frac{π}{4} s i n x - s i n \frac{π}{4} c o s x)}{x - \frac{π}{4}} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} s i n (x - \frac{π}{4})}{x - \frac{π}{4}} [? s i n (a - b) = s i n a c o s b - c o s a s i n b] \\ = \sqrt[]{2} \underset{x \to \frac{π}{4}}{l i m} \frac{s i n (x - \frac{π}{4})}{x - \frac{π}{4}} = \sqrt[]{2} . 1 = \sqrt[]{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following
$l i m_{x \to \frac{π}{6}} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{6}}{l i m} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [\frac{\sqrt[]{3}}{2} s i n x - \frac{1}{2} c o s x]}{x - \frac{π}{6}} = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [c o s \frac{π}{6} s i n x - s i n \frac{π}{6} c o s x]}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 s i n (x - \frac{π}{6})}{x - \frac{π}{6}} [? s i n (A - B) = s i n A c o s B - c o s A s i n B] \\ = 2 \underset{x \to \frac{π}{6}}{l i m} \frac{s i n (x - \frac{π}{6})}{x - \frac{π}{6}} = 2.1 = 2 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Kindly consider the following
$\underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x} \\ = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x + 3 x}{2 x} \times 2 x)}{(\frac{2 x + t a n 3 x}{3 x} \times 3 x)} = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x}{2 x} + \frac{3 x}{2 x}) \times 2 x}{(\frac{2 x}{3 x} + \frac{t a n 3 x}{3 x}) \times 3 x} \\ = \frac{(\underset{2 x \to 0}{l i m} \frac{s i n 2 x}{2 x} + \frac{3}{2})}{[\frac{2}{3} + \underset{3 x \to 0}{l i m} \frac{t a n 3 x}{3 x}]} \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ = (\frac{1 + \frac{3}{2}}{\frac{2}{3} + 1}) \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{t a n x}{x} = 1] \\ = \frac{5 / 2}{5 / 3} \times \frac{2}{3} = \frac{3}{2} \times \frac{2}{3} = 1 \end{array}$

Kindly consider the following
$l i m_{x \to a} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to a}{l i m} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}} \\ = \underset{x \to a}{l i m} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}} \times \frac{\sqrt[]{x} + \sqrt[]{a}}{\sqrt[]{x} + \sqrt[]{a}} = \underset{x \to a}{l i m} \frac{(s i n x - s i n a) (\sqrt[]{x} + \sqrt[]{a})}{x - a} \\ = \underset{x \to a}{l i m} \frac{(2 c o s \frac{x + a}{2} . s i n \frac{x - a}{2}) (\sqrt[]{x} + \sqrt[]{a})}{x - a} [? s i n A - s i n B = 2 c o s \frac{A + B}{2} . s i n \frac{A - B}{2}] \\ = \underset{\frac{x - a}{2} \to 0}{l i m} (2 c o s \frac{x + a}{2} . \frac{s i n \frac{x - a}{2}}{2 \times \frac{x - a}{2}}) (\sqrt[]{x} + \sqrt[]{a}) \\ = \underset{x \to a}{l i m} c o s (\frac{x + a}{2}) (\sqrt[]{x} + \sqrt[]{a}) [? \underset{\frac{x - a}{2} \to 0}{l i m} \frac{s i n \frac{x - a}{2}}{\frac{x - a}{2}} = 1] \\ T a k i n g, w e h a v e \\ = c o s (\frac{a + a}{2}) (\sqrt[]{a} + \sqrt[]{a}) = c o s a \times 2 \sqrt[]{a} = 2 \sqrt[]{a} . c o s a \end{array}$

Kindly consider the following

$l i m_{x \to \frac{π}{6}} \frac{c o t^{3} x - 3}{c o s e c x - 2}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{6}}{l i m} \frac{c o t^{2} x - 3}{c o s e c x - 2} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{c o s e c^{2} x - 1 - 3}{c o s e c x - 2} = \underset{x \to \frac{π}{6}}{l i m} \frac{c o s e c^{2} x - 4}{c o s e c x - 2} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{(c o s e c x - 2) (c o s e c x + 2)}{(c o s e c x - 2)} = \underset{x \to \frac{π}{6}}{l i m} (c o s e c x + 2) \\ T a k i n g, w e h a v e \\ = c o s e c \frac{π}{6} + 2 = 2 + 2 = 4 \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{\sqrt[]{2} - \sqrt[]{1 + c o s x}}{s i n^{2} x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{\sqrt[]{2} - \sqrt[]{1 + c o s x}}{s i n^{2} x} \\ = \underset{x \to 0}{l i m} \frac{\sqrt[]{2} - \sqrt[]{1 + c o s x}}{s i n^{2} x} \times \frac{\sqrt[]{2} + \sqrt[]{1 + c o s x}}{\sqrt[]{2} + \sqrt[]{1 + c o s x}} \\ = \underset{x \to 0}{l i m} \frac{2 - (1 + c o s x)}{s i n^{2} x [\sqrt[]{2} + \sqrt[]{1 + c o s x}]} = \underset{x \to 0}{l i m} \frac{1 - c o s x}{s i n^{2} x [\sqrt[]{2} + \sqrt[]{1 + c o s x}]} \\ = \underset{x \to 0}{l i m} \frac{2 s i n^{2} x / 2}{{(2 s i n x / 2 c o s x / 2)}^{2}} \times \frac{1}{[\sqrt[]{2} + \sqrt[]{1 + c o s x}]} \\ = \underset{x \to 0}{l i m} \frac{2 s i n^{2} x / 2}{4 s i n^{2} x / 2 c o s^{2} x / 2} \times \frac{1}{[\sqrt[]{2} + \sqrt[]{1 + c o s x}]} \\ = \underset{x \to 0}{l i m} \frac{2}{4 c o s^{2} x / 2} \times \frac{1}{[\sqrt[]{2} + \sqrt[]{1 + c o s x}]} \\ T a k i n g, w e h a v e \\ = \frac{2}{4 c o s^{2} 0} \times \frac{1}{[\sqrt[]{2} + \sqrt[]{2}]} = \frac{1}{2} \times \frac{1}{2 \sqrt[]{2}} = \frac{1}{4 \sqrt[]{2}} \end{array}$

Kindly consider the following
$l i m_{x \to 0} \frac{s i n x - 2 s i n 3 x + s i n 5 x}{x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n x - 2 s i n 3 x + s i n 5 x}{x} \\ = \underset{x \to 0}{l i m} \frac{s i n x}{x} - \frac{2 s i n 3 x}{x} + \frac{s i n 5 x}{x} = \underset{x \to 0}{l i m} \frac{s i n x}{x} - \underset{3 x \to 0}{l i m} 2 (\frac{s i n 3 x}{3 x}) \times 3 + \underset{5 x \to 0}{l i m} (\frac{s i n 5 x}{5 x}) \times 5 \\ = 1 - 6 + 5 = 0 . \end{array}$

If $l i m_{x \to 1} \frac{x^{4} - 1}{x - 1} = l i m_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}}$ , then find the value of $k$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{4} - 1}{x - 1} = \underset{x \to k}{l i m} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \\ \Rightarrow 4 {(1)}^{4 - 1} = \underset{x \to k}{l i m} \frac{(x - k) (x^{2} + k^{2} + k x)}{(x - k) (x + k)} \\ \Rightarrow 4 = \underset{x \to k}{l i m} \frac{x^{2} + k^{2} + k x}{x + k} \Rightarrow 4 = \frac{k^{2} + k^{2} + k^{2}}{2 k} \\ \Rightarrow 4 = \frac{3 k^{2}}{2 k} \Rightarrow 4 = \frac{3}{2} k \Rightarrow k = \frac{8}{3} . \end{array}$

Differentiate each of the functions w.r.t. $x$ in Exercises 29 to 42.

$\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = (\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}) \\ ∴ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}) \\ = \frac{d}{d x} (x^{3} + x^{2} + x + \frac{1}{x}) = 3 x^{2} + 2 x + 1 - \frac{1}{x^{2}} \end{array}$

Kindly consider the following
${(x + \frac{1}{x})}^{3}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(x + \frac{1}{x})}^{3} \\ \frac{d y}{d x} = \frac{d}{d x} {(x + \frac{1}{x})}^{3} \\ = \frac{d}{d x} (x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x}) \\ = \frac{d}{d x} (x^{3} + x^{- 3} + 3 x + 3 . x^{- 1}) = 3 x^{2} - 3 x^{- 4} + 3 - 3 . x^{- 2} \\ = 3 x^{2} - \frac{3}{x^{4}} + 3 - \frac{3}{x^{2}} \end{array}$

Kindly consider the following

$(3 x + 5) (1 + t a n x)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = (3 x + 5) (1 + t a n x) \\ \frac{d y}{d x} = \frac{d}{d x} (3 x + 5) (1 + t a n x) \\ = (3 x + 5) \frac{d}{d x} (1 + t a n x) + (1 + t a n x) \frac{d}{d x} (3 x + 5) [\begin{array}{l} ? \frac{d}{d x} [f (x) . g (x)] \\ = f (x) . g' (x) + g (x) . f' (x) \end{array}] \\ = (3 x + 5) s e c^{2} x + (1 + t a n x) (3) \\ = 3 x s e c^{2} x + 5 s e c^{2} x + 3 + 3 t a n x \end{array}$

Kindly consider the following
$(s e c x - 1) (s e c x + 1)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = (s e c x - 1) (s e c x + 1) \\ = (s e c^{2} x - 1) [? a^{2} - b^{2} = (a - b) (a + b)] \\ = t a n^{2} x [? t a n^{2} x = s e c^{2} x - 1] \\ \frac{d y}{d x} = \frac{d}{d x} t a n^{2} x \\ = 2 t a n x \frac{d}{d x} t a n x = 2 t a n x . s e c^{2} x \end{array}$

Kindly consider the following

$\frac{3 x + 4}{5 x^{2} - 7 x + 9}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{3 x + 4}{5 x^{2} - 7 x + 9} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{3 x + 4}{5 x^{2} - 7 x + 9}) \\ = \frac{(5 x^{2} - 7 x + 9) \frac{d}{d x} (3 x + 4) - (3 x + 4) \frac{d}{d x} (5 x^{2} - 7 x + 9)}{{(5 x^{2} - 7 x + 9)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(5 x^{2} - 7 x + 9) (3) - (3 x + 4) (10 x - 7)}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{15 x^{2} - 21 x + 27 - 30 x^{2} + 21 x - 40 x + 28}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{- 15 x^{2} - 40 x + 55}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{55 - 40 x - 15 x^{2}}{{(5 x^{2} - 7 x + 9)}^{2}} \end{array}$

Kindly consider the following

$\frac{x^{5} - c o s x}{s i n x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{x^{5} - c o s x}{s i n x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{5} - c o s x}{s i n x}) \\ = \frac{s i n x \frac{d}{d x} (x^{5} - c o s x) - (x^{5} - c o s x) \frac{d}{d x} (s i n x)}{s i n^{2} x} [q u o t i e n t r u l e] \\ = \frac{s i n x (5 x^{4} + s i n x) - (x^{5} - c o s x) (c o s x)}{s i n^{2} x} \\ = \frac{5 x^{4} . s i n x + s i n^{2} x - x^{5} c o s x + c o s^{2} x}{s i n^{2} x} \\ = \frac{5 x^{4} . s i n x - x^{5} c o s x + (s i n^{2} x + c o s^{2} x)}{s i n^{2} x} = \frac{5 x^{4} . s i n x - x^{5} c o s x + 1}{s i n^{2} x} \end{array}$

Kindly consider the following
$\frac{x^{2} c o s \frac{π}{4}}{s i n x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{x^{2} c o s \frac{π}{4}}{s i n x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{2} c o s \frac{π}{4}}{s i n x}) \\ = c o s \frac{π}{4} . \frac{d}{d x} (\frac{x^{2}}{s i n x}) \\ = \frac{\frac{1}{\sqrt[]{2}} [s i n x \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (s i n x)]}{s i n^{2} x} [q u o t i e n t r u l e] \\ = \frac{1}{\sqrt[]{2}} . [\frac{s i n x . 2 x - x^{2} c o s x}{s i n^{2} x}] = \frac{1}{\sqrt[]{2}} [\frac{2 x}{s i n x} - \frac{x^{2} c o s x}{s i n^{2} x}] \\ = \frac{1}{\sqrt[]{2}} [2 x c o s e c x - x^{2} c o t x c o s e c x] = \frac{x}{\sqrt[]{2}} c o s e c x [2 - x c o t x] \end{array}$

Kindly consider the following

$(a x^{2} + c o t x) (p + q c o s x)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = (a x^{2} + c o t x) (p + q c o s x) \\ \frac{d y}{d x} = \frac{d}{d x} (a x^{2} + c o t x) (p + q c o s x) \\ = (a x^{2} + c o t x) \frac{d}{d x} (p + q c o s x) + (p + q c o s x) \frac{d}{d x} (a x^{2} + c o t x) [Using p r o d u c t r u l e] \\ = (a x^{2} + c o t x) (- q s i n x) + (p + q c o s x) (2 a x - c o s e c^{2} x) \\ = - q s i n x (a x^{2} + c o t x) + (p + q c o s x) (2 a x - c o s e c^{2} x) \end{array}$

Kindly consider the following

37. $\frac{a + b s i n x}{c + d c o s x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{a + b s i n x}{c + d c o s x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{a + b s i n x}{c + d c o s x}) \\ = \frac{(c + d c o s x) \frac{d}{d x} (a + b s i n x) - (a + b s i n x) \frac{d}{d x} (c + d c o s x)}{{(c + d c o s x)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(c + d c o s x) (b c o s x) - (a + b s i n x) \frac{d}{d x} (- d s i n x)}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + b d c o s^{2} x + a d s i n x + b d s i n^{2} x}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + a d s i n x + b d (s i n^{2} x + c o s^{2} x)}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + a d s i n x + b d}{{(c + d c o s x)}^{2}} \end{array}$

Kindly consider the following
${(s i n x + c o s x)}^{2}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(s i n x + c o s x)}^{2} \\ \frac{d y}{d x} = \frac{d}{d x} {(s i n x + c o s x)}^{2} \\ = 2 (s i n x + c o s x) \frac{d}{d x} (s i n x + c o s x) \\ = 2 (s i n x + c o s x) (c o s x - s i n x) \\ = 2 (c o s^{2} x - s i n^{2} x) = 2 c o s 2 x [? c o s 2 x = c o s^{2} x - s i n^{2} x] \end{array}$

Kindly consider the following

${(2 x - 7)}^{2} {(3 x + 5)}^{3}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(2 x - 7)}^{2} {(3 x + 5)}^{3} \\ \frac{d y}{d x} = \frac{d}{d x} {(2 x - 7)}^{2} {(3 x + 5)}^{3} \\ = {(2 x - 7)}^{2} \frac{d}{d x} {(3 x + 5)}^{3} + {(3 x + 5)}^{3} \frac{d}{d x} {(2 x - 7)}^{2} [Using p r o d u c t r u l e] \\ = {(2 x - 7)}^{2} . 3 {(3 x + 5)}^{2} . 3 + {(3 x + 5)}^{3} . 2 (2 x - 7) . 2 \\ = 9 {(2 x - 7)}^{2} {(3 x + 5)}^{2} + 4 {(3 x + 5)}^{3} (2 x - 7) \\ = (2 x - 7) {(3 x + 5)}^{2} [9 (2 x - 7) + 4 (3 x + 5)] \\ = (2 x - 7) {(3 x + 5)}^{2} [18 x - 63 + 12 x + 20] \\ = (2 x - 7) {(3 x + 5)}^{2} (30 x - 43) = (2 x - 7) (30 x - 43) {(3 x + 5)}^{2} \end{array}$

Kindly consider the following
$x^{2} s i n x + c o s 2 x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = x^{2} s i n x + c o s 2 x \\ \frac{d y}{d x} = \frac{d}{d x} (x^{2} s i n x + c o s 2 x) \\ = \frac{d}{d x} (x^{2} s i n x) + \frac{d}{d x} (c o s 2 x) \\ = x^{2} c o s x + s i n x . 2 x + (- 2 s i n 2 x) \\ = x^{2} c o s x + 2 x s i n x - 2 s i n 2 x \end{array}$

Kindly consider the following
$s i n^{3} x c o s^{3} x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = s i n^{3} x c o s^{3} x \\ \frac{d y}{d x} = \frac{d}{d x} (s i n^{3} x c o s^{3} x) \\ = s i n^{3} x \frac{d}{d x} (c o s^{3} x) + c o s^{3} x . \frac{d}{d x} (s i n^{3} x) [Using p r o d u c t r u l e] \\ = s i n^{3} x . 3 c o s^{2} x (- s i n x) + c o s^{3} x . 3 s i n^{2} x . c o s x \\ = - 3 s i n^{4} x . c o s^{2} x + 3 c o s^{4} x . s i n^{2} x \\ = 3 s i n^{2} x . c o s^{2} x (- s i n^{2} x + c o s^{2} x) = 3 s i n^{2} x . c o s^{2} x . c o s 2 x \\ = \frac{3}{4} . 4 s i n^{2} x . c o s^{2} x . c o s 2 x = \frac{3}{4} {(2 s i n x . c o s x)}^{2} . c o s 2 x \\ = \frac{3}{4} s i n^{2} 2 x . c o s 2 x \end{array}$

Kindly consider the following
$\frac{1}{a x^{2} + b x + c}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{1}{a x^{2} + b x + c} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{1}{a x^{2} + b x + c}) \\ = \frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - 1 . \frac{d}{d x} (a x^{2} + b x + c)}{{(a x^{2} + b x + c)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(a x^{2} + b x + c) \times 0 - (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} = \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} \end{array}$

Limits and Derivatives Long Answer Type Questions

Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle.

1. $c o s (x^{2} + 1)$

Sol:

$\begin{array}{l} L e t f (x) = c o s (x^{2} + 1) \dots (i) \\ \Rightarrow f (x + Δ x) = c o s [{(x + Δ x)}^{2} + 1] \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1) \\ D i v i d i n g b o t h s i d e s b y Δ x w e g e t \\ \frac{f (x + Δ x) - f (x)}{Δ x} = \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [\frac{{(x + Δ x)}^{2} + 1 + x^{2} + 1}{2}] . s i n [\frac{{(x + Δ x)}^{2} + 1 - x^{2} - 1}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [\frac{x^{2} + Δ x^{2} + 2 x Δ x + x^{2} + 2}{2}] . s i n [\frac{x^{2} + Δ x^{2} + 2 x Δ x - x^{2}}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] . s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] . s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x [\frac{Δ x + 2 x}{2}]} \times [\frac{Δ x + 2 x}{2}] \\ = \underset{Δ x [\frac{Δ x + 2 x}{2}] \to 0}{l i m} - 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] \frac{s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x [\frac{Δ x + 2 x}{2}]} \times [\frac{Δ x + 2 x}{2}] \\ T a k i n g limit, w e h a v e \\ = - 2 s i n (x^{2} + 1) . 1 . (x) = - 2 x s i n (x^{2} + 1) . \end{array}$

2. $\frac{a x + b}{c x + d}$

Sol:

$\begin{array}{l} L e t f (x) = \frac{a x + b}{c x + d} \dots (i) \\ \Rightarrow f (x + Δ x) = \frac{a (x + Δ x) + b}{c (x + Δ x) + d} \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = \frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d} \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{\frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d}}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{(a x + a Δ x + b) (c x + d) - (a x + b) (c x + c Δ x + d)}{[c (x + Δ x) + d] (c x + d) . Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{a c x^{2} + a c Δ x . x + b c x + a d x + a d Δ x + b d - a c x^{2} - a c Δ x . x - a d x - b c x - b c . Δ x - b d}{(c x + c Δ x + d) (c x + d) Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{(a d - b c) Δ x}{(c x + c Δ x + d) (c x + d) . Δ x} = \underset{Δ x \to 0}{l i m} \frac{(a d - b c)}{(c x + c . Δ x + d) (c x + d)} \\ T a k i n g limit, w e h a v e \\ = \frac{(a d - b c)}{(c x + d) (c x + d)} = \frac{(a d - b c)}{{(c x + d)}^{2}} \end{array}$

Commonly asked questions

Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle.

$c o s (x^{2} + 1)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$\frac{a x + b}{c x + d}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$x^{\frac{2}{3}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = x^{2 / 3} \dots (i) \\ \Rightarrow f (x + Δ x) = {(x + Δ x)}^{2 / 3} \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = {(x + Δ x)}^{2 / 3} - x^{2 / 3} \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{{(x + Δ x)}^{2 / 3} - x^{2 / 3}}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} {[1 + \frac{Δ x}{x}]}^{2 / 3} - x^{2 / 3}}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} [{(1 + \frac{Δ x}{x})}^{2 / 3} - 1]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} [(1 + \frac{2}{3} . \frac{Δ x}{x} + \dots) - 1]}{Δ x} [\begin{array}{l} E x p a n d i n g b y B i n o m i a l t h e o r e m a n d r e j e c t i n g \\ t h e h i g h e r p o w e r s o f Δ x a s Δ x \to 0 \end{array}] \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} . \frac{2}{3} . \frac{Δ x}{x}}{Δ x} = \frac{2}{3} x^{2 / 3 - 1} = \frac{2}{3} x^{- 1 / 3} . \end{array}$

Kindly consider the following

$x c o s x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = x c o s x \dots (i) \\ \Rightarrow y + Δ y = (x + Δ x) c o s (x + Δ x) \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ y + Δ y - y = (x + Δ x) c o s (x + Δ x) - x c o s x \\ \Rightarrow Δ y = x c o s (x + Δ x) + Δ x c o s (x + Δ x) - x c o s x \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e limit w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{Δ y}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{x c o s (x + Δ x) - x c o s x + Δ x c o s (x + Δ x)}{Δ x} \\ \frac{d y}{d x} = \underset{Δ x \to 0}{l i m} \frac{x [c o s (x + Δ x) - c o s x]}{Δ x} + \underset{Δ x \to 0}{l i m} \frac{Δ x c o s (x + Δ x)}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x [- 2 s i n \frac{(x + Δ x + x)}{2} . s i n \frac{(x + Δ x - x)}{2}]}{Δ x} + \underset{Δ x \to 0}{l i m} c o s (x + Δ x) \\ = \underset{\begin{array}{l} Δ x \to 0 \\ ∴ \frac{Δ x}{2} \to 0 \end{array}}{l i m} \frac{x [- 2 s i n (x + \frac{Δ x}{2}) . s i n \frac{Δ x}{2}]}{2 \times \frac{Δ x}{2}} + \underset{Δ x \to 0}{l i m} c o s (x + Δ x) \\ ∴ \frac{Δ x}{2} \to 0 T a k i n g limits w e h a v e \\ = x [- s i n x] + c o s x [? \underset{\frac{Δ x}{2} \to 0}{l i m} \frac{s i n \frac{Δ x}{2}}{\frac{Δ x}{2}} = 1] \\ = - x s i n x + c o s x \end{array}$

Evaluate each of the following limits in Exercises 47 to 53.

$l i m_{y \to 0} \frac{(x + y) s e c (x + y) - x s e c x}{y^{}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{y \to 0}{l i m} \frac{(x + y) s e c (x + y) - x s e c x}{y} \\ = \underset{y \to 0}{l i m} \frac{x s e c (x + y) + y s e c (x + y) - x s e c x}{y} \\ = \underset{y \to 0}{l i m} \frac{[x s e c (x + y) - x s e c x]}{y} + \underset{y \to 0}{l i m} \frac{y s e c (x + y)}{y} \\ = \underset{y \to 0}{l i m} \frac{x [s e c (x + y) - s e c x]}{y} + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} \frac{x [\frac{1}{c o s (x + y)} - \frac{1}{c o s x}]}{y} + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{c o s x - c o s (x + y)}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{- 2 s i n (\frac{x + x + y}{2}) . s i n (\frac{x - x - y}{2})}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{- 2 s i n (x + \frac{y}{2}) . s i n (\frac{- y}{2})}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{\begin{array}{l} y \to 0 \\ ∴ \frac{y}{2} \to 0 \end{array}}{l i m} x [\frac{2 s i n (x + \frac{y}{2}) . s i n (\frac{y}{2})}{c o s x . c o s (x + y) . (\frac{y}{2}) . 2}] + \underset{y \to 0}{l i m} s e c (x + y) \\ T a k i n g w e h a v e \\ = x [s i n x . \frac{1}{c o s x . c o s x}] + s e c x \\ = x s e c x t a n x + s e c x = s e c x (x t a n x + 1) \end{array}$

Kindly consider the following
$\underset{x \to 0}{l i m} \frac{s i n (α + β) x - s i n (α - β) x + s i n 2 α x}{c o s^{2} β x - c o s^{2} α x} . x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n (α + β) x - s i n (α - β) x + s i n 2 α x}{c o s^{2} β x - c o s^{2} α x} . x \\ = \underset{x \to 0}{l i m} \frac{[2 s i n α x . c o s β x + s i n 2 α x] . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{[2 s i n α x . c o s β x + 2 s i n α x . c o s α x] . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{2 s i n α x (c o s β x + c o s α x) . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x [2 c o s (\frac{α + β}{2}) x . c o s (\frac{α - β}{2}) x] . x}{s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x [2 c o s (\frac{α + β}{2}) x . c o s (\frac{α - β}{2}) x] . x}{2 s i n (\frac{α + β}{2}) x . c o s (\frac{α + β}{2}) x . 2 s i n (\frac{α - β}{2}) x . c o s (\frac{α - β}{2}) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x . x}{2 s i n (\frac{α + β}{2}) x s i n (\frac{α - β}{2}) x} \\ = \underset{x \to 0}{l i m} \frac{1}{2} \frac{\frac{s i n α x}{α x} . (α x) . x}{[\frac{s i n (\frac{α + β}{2}) x}{(\frac{α + β}{2}) x} \times (\frac{α + β}{2}) . x] [\frac{s i n (\frac{α - β}{2}) x}{(\frac{α - β}{2}) x} \times (\frac{α - β}{2}) . x]} \\ = \frac{1}{2} . \frac{α x^{2}}{(\frac{α + β}{2}) x (\frac{α - β}{2}) x} = \frac{1}{2} . [\frac{α}{(\frac{α + β}{2}) (\frac{α - β}{2})}] \\ = \frac{1}{2} . \frac{4 α}{α^{2} - β^{2}} = \frac{2 α}{α^{2} - β^{2}} \end{array}$

Kindly consider the following

$\underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{3} x - t a n x}{c o s (x + \frac{π}{4})}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{3} x - t a n x}{c o s (x + \frac{π}{4})} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{t a n x (t a n^{2} x - 1)}{c o s (x + \frac{π}{4})} = \underset{x \to \frac{π}{4}}{l i m} t a n x . \underset{x \to \frac{π}{4}}{l i m} [\frac{- (1 - t a n^{2} x)}{c o s (x + \frac{π}{4})}] \\ = - 1 \times \underset{x \to \frac{π}{4}}{l i m} \frac{(1 - t a n x) (1 + t a n x)}{c o s (x + \frac{π}{4})} \\ = \underset{x \to \frac{π}{4}}{l i m} - (1 + t a n x) \times \underset{x \to \frac{π}{4}}{l i m} [\frac{1 - t a n x}{c o s (x + \frac{π}{4})}] \\ = - (1 + 1) \times \underset{x \to \frac{π}{4}}{l i m} \frac{(c o s x - s i n x)}{c o s x . c o s (x + \frac{π}{4})} = - 2 \times \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (\frac{1}{\sqrt[]{2}} c o s x - \frac{1}{\sqrt[]{2}} s i n x)}{c o s x . c o s (x + \frac{π}{4})} \\ = - 2 \sqrt[]{2} \times \underset{x \to \frac{π}{4}}{l i m} \frac{(c o s \frac{π}{4} . c o s x - s i n \frac{π}{4} . s i n x)}{c o s x . c o s (x + \frac{π}{4})} \\ = - 2 \sqrt[]{2} \times \underset{x \to \frac{π}{4}}{l i m} \frac{c o s (x + \frac{π}{4})}{c o s x . c o s (x + \frac{π}{4})} = - 2 \sqrt[]{2} \times \underset{x \to \frac{π}{4}}{l i m} \frac{1}{c o s x} \\ T a k i n g limit w e h a v e \\ = \frac{- 2 \sqrt[]{2}}{c o s \frac{π}{4}} = \frac{- 2 \sqrt[]{2}}{\frac{1}{\sqrt[]{2}}} = - 2 \times 2 = - 4 . \end{array}$

Kindly consider the following
$\underset{x \to π}{l i m} \frac{1 - s i n \frac{x}{2}}{c o s \frac{x}{2} (c o s \frac{x}{4} - s i n \frac{x}{4})}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{1 - s i n \frac{x}{2}}{c o s \frac{x}{2} (c o s \frac{x}{4} - s i n \frac{x}{4})} \\ = \underset{x \to π}{l i m} \frac{c o s^{2} \frac{x}{4} + s i n^{2} \frac{x}{4} - 2 s i n \frac{x}{4} . c o s \frac{x}{4}}{(c o s^{2} \frac{x}{4} - s i n^{2} \frac{x}{4}) (c o s \frac{x}{4} - s i n \frac{x}{4})} [\begin{array}{l} ? c o s 2 x = c o s^{2} x - s i n^{2} x \\ s i n^{2} x + c o s^{2} x = 1 \end{array}] \\ = \underset{x \to π}{l i m} \frac{{(c o s \frac{x}{4} - s i n \frac{x}{4})}^{2}}{(c o s \frac{x}{4} - s i n \frac{x}{4}) (c o s \frac{x}{4} + s i n \frac{x}{4}) (c o s \frac{x}{4} - s i n \frac{x}{4})} \\ = \underset{x \to π}{l i m} \frac{1}{(c o s \frac{x}{4} + s i n \frac{x}{4})} \\ T a k i n g limit w e h a v e \\ = \frac{1}{c o s \frac{π}{4} + s i n \frac{π}{4}} = \frac{1}{\frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{2}}} = \frac{1}{\frac{2}{\sqrt[]{2}}} = \frac{1}{\sqrt[]{2}} . \end{array}$

Show that $l i m_{x \to 4} \frac{? x - 4 ?}{x - 4}$ does not exist.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n \underset{x \to 4}{l i m} \frac{| x - 4 |}{x - 4} \\ L H L = \underset{x \to 4^{-}}{l i m} \frac{- (x - 4)}{x - 4} = - 1 [? | x - 4 | = - (x - 4) i f x < 4] \\ R H L = \underset{x \to 4^{+}}{l i m} \frac{(x - 4)}{x - 4} = 1 [? | x - 4 | = (x - 4) i f x > 4] \\ L H L \neq R H L \\ H e n c e, t h e limit d o e s n o t e x i s t . \end{array}$

Let $f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & when x \neq \frac{π}{2}, \\ 3 & when x = \frac{π}{2} \end{matrix}$ , and if $l i m_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of $k$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & w h e n x \neq \frac{π}{2}, \\ 3 & w h e n x = \frac{π}{2} \end{matrix} \\ L H L f (x) = \underset{x \to {\frac{π}{2}}^{-}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \\ = \underset{h \to 0}{l i m} \frac{k s i n h}{π - π + 2 h} = \underset{h \to 0}{l i m} \frac{k s i n h}{2 h} = \frac{k}{2} . 1 = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L f (x) = \underset{x \to {\frac{π}{2}}^{+}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \underset{h \to 0}{l i m} \frac{- k s i n h}{π - π - 2 h} = \underset{h \to 0}{l i m} \frac{- k s i n h}{- 2 h} = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ W e a r e g i v e n t h a t \underset{x \to \frac{π}{2}}{l i m} f (x) = 3 \\ S o, \frac{k}{2} = 3 \Rightarrow k = 6 \end{array}$

Let $f (x) = {\begin{matrix} x^{} + 2 & if x \leq - 1, \\ c x^{2} & if x > - 1 \end{matrix}$ , find ‘ $c$ ’ if $l i m_{x \to - 1} f (x)$ exists

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} x + 2 & i f x \leq - 1, \\ c x^{2} & i f x > - 1 \end{matrix} \\ L H L f (x) = \underset{x \to - 1^{-}}{l i m} (x + 2) = \underset{h \to 0}{l i m} (- 1 - h + 2) \\ = \underset{h \to 0}{l i m} (1 - h) = 1 \\ R H L f (x) = \underset{x \to - 1^{+}}{l i m} c x^{2} = \underset{h \to 0}{l i m} c {(- 1 + h)}^{2} = c \\ Since t h e limits e x i t . \\ ∴ L H L = R H L \\ ∴ c = 1 \end{array}$

Limits and Derivatives Objective Type Questions

1. Choose the correct answer from the given four options in each of the Exercises 54 to 76:

$l i m_{x \to π} \frac{s i n x}{x - π}$ is

(a) 1

(b) 2

(c) -1

(d) 2

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{s i n x}{x - π} \\ = \underset{x \to π}{l i m} \frac{s i n (π - x)}{- (π - x)} = - 1 [∵ \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1 a n d π - x \to 0 \Rightarrow x \to π] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

2. $l i m_{x \to 0} \frac{x^{2} c o s x}{1 - c o s x}$ is

(a) 2

(b) $\frac{3}{2}$
(c) $- \frac{3}{2}$
(d) 1

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{1 - c o s x} \\ = \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{2 s i n^{2} \frac{x}{2}} = - 1 [∵ 1 - c o s x = 2 s i n^{2} \frac{x}{2}] \\ = \underset{x \to 0}{l i m} \frac{\frac{x^{2}}{4} \times 4 c o s x}{2 s i n^{2} \frac{x}{2}} = \underset{\begin{array}{l} x \to 0 \\ \frac{x}{2} \to 0 \end{array}}{l i m} \frac{{(\frac{x}{2})}^{2} \times 2 c o s x}{s i n^{2} \frac{x}{2}} \\ = \underset{\frac{x}{2} \to 0}{l i m} {(\frac{\frac{x}{2}}{s i n \frac{x}{2}})}^{2} \times 2 c o s x = 2 c o s 0 = 2 \times 1 = 2 [∵ \underset{x \to 0}{l i m} \frac{x}{s i n x} = 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Commonly asked questions

Choose the correct answer from the given four options in each of the Exercises 54 to 76:

$l i m_{x \to π} \frac{s i n x}{x - π}$ is

(a) 1

(b) 2

(c) -1

(d) 2

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{s i n x}{x - π} \\ = \underset{x \to π}{l i m} \frac{s i n (π - x)}{- (π - x)} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1 a n d π - x \to 0 \Rightarrow x \to π] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

$l i m_{x \to 0} \frac{x^{2} c o s x}{1 - c o s x}$ is

(a) 2

(b) $\frac{3}{2}$
(c) $- \frac{3}{2}$
(d) 1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{1 - c o s x} \\ = \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{2 s i n^{2} \frac{x}{2}} = - 1 [? 1 - c o s x = 2 s i n^{2} \frac{x}{2}] \\ = \underset{x \to 0}{l i m} \frac{\frac{x^{2}}{4} \times 4 c o s x}{2 s i n^{2} \frac{x}{2}} = \underset{\begin{array}{l} x \to 0 \\ \frac{x}{2} \to 0 \end{array}}{l i m} \frac{{(\frac{x}{2})}^{2} \times 2 c o s x}{s i n^{2} \frac{x}{2}} \\ = \underset{\frac{x}{2} \to 0}{l i m} {(\frac{\frac{x}{2}}{s i n \frac{x}{2}})}^{2} \times 2 c o s x = 2 c o s 0 = 2 \times 1 = 2 [? \underset{x \to 0}{l i m} \frac{x}{s i n x} = 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Kindly consider the following
$l i m_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is

(a) 1
(b) $\frac{m}{n}$
(c) $\frac{m}{n} - 1$
(d) $\frac{m^{2}}{n^{2}}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{m} - 1}{x^{n} - 1} \\ = \underset{x \to 1}{l i m} \frac{\frac{x^{m} - {(1)}^{m}}{x - 1}}{\frac{x^{n} - {(1)}^{n}}{x - 1}} = \frac{m {(1)}^{m - 1}}{n {(1)}^{n - 1}} = \frac{m}{n} [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Kindly consider the following

$l i m_{θ \to 0} \frac{1 - c o s 4 θ}{1 - c o s 6 θ}$ is

(a) $\frac{4}{9}$
(b) $\frac{1}{2}$
(c) $- \frac{1}{2}$
(d) -1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{θ \to 0}{l i m} \frac{1 - c o s 4 θ}{1 - c o s 6 θ} \\ = \underset{θ \to 0}{l i m} \frac{2 s i n^{2} 2 θ}{2 s i n^{2} 3 θ} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{θ \to 0}{l i m} \frac{s i n^{2} 2 θ}{s i n^{2} 3 θ} = \underset{θ \to 0}{l i m} {[\frac{s i n 2 θ}{s i n 3 θ}]}^{2} = \underset{\begin{array}{l} θ \to 0 \\ 2 θ \to 0 \\ 3 θ \to 0 \end{array}}{l i m} {[\frac{\frac{s i n 2 θ}{2 θ} \times 2 θ}{\frac{s i n 3 θ}{3 θ} \times 3 θ}]}^{2} \\ = {[\frac{2 θ}{3 θ}]}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Kindly consider the following
$l i m_{x \to 0} \frac{c o s e c x - c o t x}{x}$ is

(a) $- \frac{1}{2}$
(b) 1
(c) $\frac{1}{2}$
(d) 1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{c o s e c x - c o t x}{x} \\ = \underset{x \to 0}{l i m} \frac{\frac{1}{s i n x} - \frac{c o s x}{s i n x}}{x} = \underset{x \to 0}{l i m} \frac{1 - c o s x}{x s i n x} = \underset{x \to 0}{l i m} \frac{2 s i n^{2} \frac{x}{2}}{x . 2 s i n \frac{x}{2} c o s \frac{x}{2}} \\ = \underset{x \to 0}{l i m} \frac{s i n \frac{x}{2}}{x c o s \frac{x}{2}} = \underset{x \to 0}{l i m} \frac{t a n \frac{x}{2}}{x} = \underset{x \to 0}{l i m} \frac{t a n \frac{x}{2}}{2 \times \frac{x}{2}} = \frac{1}{2} \times 1 = \frac{1}{2} [? \underset{x \to 0}{l i m} \frac{t a n x}{x} = 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}}$ is

(a) 2
(b) 0
(c) 1
(d) -1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} \times \frac{\sqrt[]{x + 1} + \sqrt[]{1 - x}}{\sqrt[]{x + 1} + \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{x + 1 - 1 + x} = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{2 x} \\ = \frac{1}{2} . \underset{x \to 0}{l i m} \frac{s i n x}{x} [\sqrt[]{x + 1} + \sqrt[]{1 - x}] \\ T a k i n g limit, w e g e t \\ = \frac{1}{2} \times 1 \times [\sqrt[]{0 + 1} + \sqrt[]{1 - 0}] = \frac{1}{2} \times 1 \times 2 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Kindly consider the following

$\underset{x \to \frac{π}{4}}{l i m} \frac{s e c^{2} x - 2}{t a n x - 1}$ is

(a) 3
(b) 1
(c) 0
(d) √2

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{s e c^{2} x - 2}{t a n x - 1} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{1 + t a n^{2} x - 2}{t a n x - 1} = \underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{2} x - 1}{t a n x - 1} = \underset{x \to \frac{π}{4}}{l i m} \frac{(t a n x + 1) (t a n x - 1)}{(t a n x - 1)} \\ = \underset{x \to \frac{π}{4}}{l i m} (t a n x + 1) = t a n \frac{π}{4} + 1 = 1 + 1 = 2 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Kindly consider the following
$\underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3}$ is

(a) $\frac{1}{10}$
(b) $- \frac{1}{10}$
(c) 1
(d) None of these

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + 3 x - 2 x - 3} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{x (2 x + 3) - 1 (2 x + 3)} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{(2 x + 3) (x - 1)} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (\sqrt[]{x} + 1) (2 x - 3)}{(2 x + 3) (x - 1) (\sqrt[]{x} + 1)} \\ = \underset{x \to 1}{l i m} \frac{(x - 1) (2 x - 3)}{(x - 1) (\sqrt[]{x} + 1) (2 x + 3)} = \underset{x \to 1}{l i m} \frac{2 x - 3}{(\sqrt[]{x} + 1) (2 x + 3)} \\ T a k i n g limits w e h a v e \\ = \frac{2 (1) - 3}{(\sqrt[]{1} + 1) (2 \times 1 + 3)} = \frac{- 1}{2 \times 5} = \frac{- 1}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & if [x] \neq 0, \\ 0, & [x] = 0 \end{matrix}$ , where [.] denotes the greatest integer function, then $l i m_{x \to 0} f (x)$ is equal to

(a) 1

(b) 0

(c) –1

(d) None of these

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & i f [x] \neq 0, \\ 0, & [x] = 0 \end{matrix} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 - h]}{[0 - h]} = \underset{h \to 0}{l i m} \frac{- s i n [- h]}{[- h]} = - 1 \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 + h]}{[0 + h]} = \underset{h \to 0}{l i m} \frac{s i n [h]}{[h]} = - 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Kindly consider the following
$l i m_{x \to 0} \frac{∣ s i n x ∣}{x}$ is

(a) 1

(b) –1

(c) Does not exist

(d) None of these

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{| s i n x |}{x} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{- s i n x}{x} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n x}{x} = 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Let $f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2, \\ 2 x + 3, & 2 \leq x < 3 \end{matrix}$ , the quadratic equation whose roots are $l i m_{x \to 2^{-}} f (x)$ and $l i m_{x \to 2^{+}} f (x)$ is

(a) $x^{2} - 6 x + 9 = 0$
(b) $x^{2} - 7 x + 8 = 0$
(c) $x^{2} - 14 x + 49 = 0$
(d) $x^{2} - 10 x + 21 = 0$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{matrix} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} (x^{2} - 1) = \underset{h \to 0}{l i m} [{(2 - h)}^{2} - 1] = \underset{h \to 0}{l i m} (4 + h^{2} - 4 h - 1) \\ = \underset{h \to 0}{l i m} (h^{2} - 4 h + 3) = 3 \\ a n d \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} (2 x + 3) = \underset{h \to 0}{l i m} [2 (2 + h) + 3] = 7 \\ T h e r e f o r e, t h e q u a d r a t i c e q u a t i o n w h o s e r o o t s a r e 3 a n d 7 i s \\ x^{2} - (3 + 7) x + 3 \times 7 = 0 i . e ., x^{2} - 10 x + 21 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Kindly consider the following

$l i m_{x \to 0} \frac{t a n 2 x - x}{3 x - s i n x}$ is

(a) 2

(b) $\frac{1}{2}$

(c) $- \frac{1}{2}$

(d) $\frac{1}{4}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{t a n 2 x - x}{3 x - s i n x} \\ = \underset{x \to 0}{l i m} \frac{x [\frac{t a n 2 x}{x} - 1]}{x [3 - \frac{s i n x}{x}]} = \underset{\begin{array}{l} x \to 0 \\ ∴ 2 x \to 0 \end{array}}{l i m} \frac{\frac{t a n 2 x}{2 x} \times 2 - 1}{3 - \frac{s i n x}{x}} \\ = \frac{1.2 - 1}{3 - 1} = \frac{2 - 1}{2} = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let $f (x) = x - [x]$ ; $x \in ℝ$

, then $f^{'} (\frac{1}{2})$ i

(a) $\frac{3}{2}$

(b) 1

(c) 0

(d) –1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = x - [x] \\ W e h a v e t o f i r s t c h e c k d i f f e r e n t i a b i l i t y o f f (x) a t x = \frac{1}{2} \\ ∴ L f' (\frac{1}{2}) = L H D = \underset{h \to 0}{l i m} \frac{f [\frac{1}{2} - h] - f [\frac{1}{2}]}{- h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} - h) - [\frac{1}{2} - h] - \frac{1}{2} + [\frac{1}{2}]}{- h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} - h - 0 - \frac{1}{2} + 0}{- h} = \frac{- h}{- h} = 1 \\ L f' (\frac{1}{2}) = R H D = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} + h) - [\frac{1}{2} + h] - \frac{1}{2} + [\frac{1}{2}]}{h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} + h - 1 - \frac{1}{2} + 1}{h} = \frac{h}{h} = 1 \\ L H D = R H D \\ ∴ f' (\frac{1}{2}) = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Kindly consider the following

If $y = \sqrt[]{x} + \frac{1}{\sqrt x}$ , then $\frac{d y}{d x}$ at $x = 1$ is

(a) 1

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt 2}$

(d) 0

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \sqrt[]{x} + \frac{1}{\sqrt[]{x}} \\ \frac{d y}{d x} = \frac{1}{2 \sqrt[]{x}} - \frac{1}{2 x^{3 / 2}} \\ {(\frac{d y}{d x})}_{a t x = 1} = \frac{1}{2} - \frac{1}{2} = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Kindly consider the following

If $f (x) = \frac{x - 4}{2 \sqrt x}$ , then $f^{'} (1)$ is

(a) $\frac{5}{4}$

(b) $\frac{4}{5}$

(c) 1

(d) 0

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{x - 4}{2 \sqrt[]{x}} \\ ∴ f' (x) = \frac{1}{2} [\frac{\sqrt[]{x} . 1 - (x - 4) . \frac{1}{2 \sqrt[]{x}}}{x}] \\ = \frac{1}{2} [\frac{2 x - x + 4}{2 \sqrt[]{x} . x}] = \frac{1}{2} [\frac{x + 4}{2 {(x)}^{3 / 2}}] \\ ∴ f' (x) a t x = 1 = \frac{1}{2} [\frac{1 + 4}{2 \times 1}] = \frac{5}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Kindly consider the following

If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}}$ , then $\frac{d y}{d x}$ is

(a) $\frac{- 4 x}{(x^{2} - 1)^{2}}$
(b) $\frac{- 4 x}{x^{2} - 1}$
(c) $\frac{1 - x^{2}}{4 x}$
(d) $\frac{4}{x^{2} - 1}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} \\ \Rightarrow y = \frac{x^{2} + 1}{x^{2} - 1} \\ ∴ \frac{d y}{d x} = \frac{(x^{2} - 1) . 2 x - (x^{2} + 1) . 2 x}{{(x^{2} - 1)}^{2}} \\ = \frac{2 x (x^{2} - 1 - x^{2} - 1)}{{(x^{2} - 1)}^{2}} = \frac{2 x (- 2)}{{(x^{2} - 1)}^{2}} = \frac{- 4 x}{{(x^{2} - 1)}^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $y = \frac{s i n x + c o s x}{s i n x - c o s x}$ , then $\frac{d y}{d x}$ at $x = 0$ is

(a) –2

(b) 0

(c) $\frac{1}{2}$

(d) Does not exist

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \frac{s i n x + c o s x}{s i n x - c o s x} \\ ∴ \frac{d y}{d x} = \frac{(s i n x - c o s x) (c o s x - s i n x) - (s i n x + c o s x) (c o s x + s i n x)}{{(s i n x - c o s x)}^{2}} \\ = \frac{- {(s i n x - c o s x)}^{2} - {(s i n x + c o s x)}^{2}}{{(s i n x - c o s x)}^{2}} \\ = \frac{- [s i n^{2} x + c o s^{2} x - 2 s i n x c o s x + s i n^{2} x + c o s^{2} x + 2 s i n x c o s x]}{{(s i n x - c o s x)}^{2}} = \frac{- 2}{{(s i n x - c o s x)}^{2}} \\ ∴ {(\frac{d y}{d x})}_{a t x = 0} = \frac{- 2}{{(s i n 0 - c o s 0)}^{2}} = \frac{- 2}{{(- 1)}^{2}} = - 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $y = \frac{s i n (x + 9)}{c o s x}$ , then $\frac{d y}{d x}$ at $x = 0$ is

(a) cos9

(b) sin9

(c) 0

(d) 1

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \frac{s i n (x + 9)}{c o s x} \\ ∴ \frac{d y}{d x} = \frac{c o s x . c o s (x + 9) - s i n (x + 9) (- s i n x)}{c o s^{2} x} \\ = \frac{c o s x c o s (x + 9) + s i n x s i n (x + 9)}{c o s^{2} x} \\ = \frac{c o s (x + 9 - x)}{c o s^{2} x} = \frac{c o s 9}{c o s^{2} x} \\ ∴ {(\frac{d y}{d x})}_{a t x = 0} = \frac{c o s 9}{c o s^{2} 0} = \frac{c o s 9}{{(1)}^{2}} = c o s 9 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100}$ , then $f^{'} (1)$ is equal to

(a) $\frac{1}{100}$
(b) 100
(c) Does not exist
(d) 0

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100} \\ f' (x) = 1 + \frac{2 x}{2} + \dots + \frac{100 x^{99}}{100} \\ ∴ f' (1) = 1 + 1 + 1 + \dots + 1 (100 t i m e s) = 100 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant $a$ , then $f^{'} (a)$ is

(a) 1

(b) 0

(c) Does not exist

(d) $\frac{1}{2}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{x^{n} - a^{n}}{x - a} \\ f' (x) = \frac{(x - a) (n . x^{n - 1}) - (x^{n} - a^{n}) . 1}{{(x - a)}^{2}} \\ ∴ f' (a) = \frac{(a - a) (n . a^{n - 1}) - (a^{n} - a^{n}) . 1}{{(a - a)}^{2}} \\ S o, f' (a) = \frac{0}{0} = d o e s n o t e x i s t \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $f (x) = x^{100} + x^{99} + . . . + x + 1$ , then $f^{'} (1)$ is equal to

(a) 5050

(b) 5049

(c) 5051

(d) 5005

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{100} + x^{99} + \dots + x + 1 \\ ∴ f' (x) = 100 x^{99} + 99 x^{98} + \dots + 1 \\ S o, f' (1) = 100 + 99 + 98 + \dots + 1 \\ = \frac{100}{2} [2 \times 100 + (100 - 1) (- 1)] = 50 [200 - 99] = 50 \times 101 = 5050 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $f (x) = 1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100}$ , then $f^{'} (1)$ is equal to

(a) 150

(b) –50

(c) –150

(d) 50

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = 1 - x + x^{2} - x^{3} + \dots - x^{99} + x^{100} \\ ∴ f' (x) = - 1 + 2 x - 3 x^{2} + \dots - 99 x^{98} + 100 x^{99} \\ S o, f' (1) = - 1 + 2 - 3 + \dots - 99 + 100 \\ = (- 1 - 3 - 5 \dots - 99) + (2 + 4 + 6 + \dots + 100) \\ = \frac{50}{2} [2 \times - 1 + (50 - 1) (- 2)] + \frac{50}{2} [2 \times 2 + (50 - 1) (2)] \\ = 25 [- 2 - 98] + 25 [4 + 98] = 25 \times - 100 + 25 \times 102 \\ = 25 [- 100 + 102] = 25 \times 2 = 50 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x}$ is

(a) $n$

(b) 1

(c) – $n$

(d) 0

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x} \\ = \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = \underset{1 + x \to 1}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = n {(1)}^{n - 1} = n [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Limits and Derivatives Fill in the blanks

1. If $f (x) = \frac{t a n x}{x - π}$ , then $l i m_{x \to π} f (x) =_________$ .

Sol:

$\begin{array}{l} G i v e n f (x) = \underset{x \to π}{l i m} \frac{- t a n (π - x)}{x - π} \\ = \underset{π - x \to 0}{l i m} \frac{- t a n (π - x)}{- (π - x)} = 1 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 1 . \end{array}$

2. $l i m_{x \to 0} s i n (m x) c o t$ x/ $\sqrt$ 3 = 2 then m = __________.

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} (s i n m x c o t \frac{x}{\sqrt[]{3}}) = 2 \\ \Rightarrow \underset{\begin{array}{l} x \to 0 \\ ∴ m x \to 0 \end{array}}{l i m} \frac{s i n m x}{m x} \times m x \underset{x \to 0}{l i m} (c o t \frac{x}{\sqrt[]{3}}) = 2 \\ \Rightarrow 1 \times m x \underset{x \to 0}{l i m} \frac{1}{t a n \frac{x}{\sqrt[]{3}}} = 2 \\ \Rightarrow \underset{x \to 0}{l i m} m x \times \frac{\frac{x}{\sqrt[]{3}}}{\frac{x}{\sqrt[]{3}} . t a n \frac{x}{\sqrt[]{3}}} = 2 \\ \Rightarrow \frac{m x}{\frac{x}{\sqrt[]{3}}} \times (1) = 2 \Rightarrow \sqrt[]{3} m = 2 \Rightarrow m = \frac{2}{\sqrt[]{3}} = \frac{2 \sqrt[]{3}}{3} . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{2 \sqrt[]{3}}{3} . \end{array}$

Commonly asked questions

If $f (x) = \frac{t a n x}{x - π}$ , then $l i m_{x \to π} f (x) =_________$ .

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$l i m_{x \to 0} s i n (m x) c o t$ x/ $\sqrt$ 3 = 2 then m = __________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$ , then $\frac{d y}{d x} =_________.$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots \\ \frac{d y}{d x} = 0 + \frac{1}{1!} + \frac{2 x}{2!} + \frac{3 x^{2}}{3!} + \dots \\ = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = y \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s y . \end{array}$

Kindly consider the following
$l i m_{x \to 3} \frac{x}{[x]} =_________$ .

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n \underset{x \to 3^{+}}{l i m} \frac{x}{[x]} \\ = \underset{h \to 0}{l i m} \frac{(3 + h)}{[3 + h]} = 1 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 1 . \end{array}$

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exam

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Limits and Derivatives

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Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen: Overview, Questions, Preparation

Limits and Derivatives Short Answer Type Questions

Commonly asked questions

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Commonly asked questions

Limits and Derivatives Objective Type Questions

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Limits and Derivatives Fill in the blanks

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