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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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) $\frac{m^{2}}{n^{2}}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

(a) 2

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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