Limits and Derivatives

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
=

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-1}{x}\\ =\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=\underset{1+x\to 1}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=n{\left(1\right)}^{n-1}=n\left[?\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n.a^{n-1}\right]\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

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

Sol:

$\begin{array}{l}Given\underset{x\to 3^{+}}{lim}\frac{x}{\left[x\right]}\\ =\underset{h\to 0}{lim}\frac{\left(3+h\right)}{\left[3+h\right]}=1\\ Hence,thevalueofthefilleris1.\end{array}$

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

Sol:

$\begin{array}{l}Giventhaty=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ \frac{dy}{dx}=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\dots \\ =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots =y\\ Hence,thevalueofthefillerisy.\end{array}$

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

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\left(sinmxcot\frac{x}{\sqrt[]{3}}\right)=2\\ \Rightarrow \underset{\begin{array}{l}x\to 0\\ \therefore mx\to 0\end{array}}{lim}\frac{sinmx}{mx}\times mx\underset{x\to 0}{lim}\left(cot\frac{x}{\sqrt[]{3}}\right)=2\\ \Rightarrow 1\times mx\underset{x\to 0}{lim}\frac{1}{tan\frac{x}{\sqrt[]{3}}}=2\\ \Rightarrow \underset{x\to 0}{lim}mx\times \frac{\frac{x}{\sqrt[]{3}}}{\frac{x}{\sqrt[]{3}}.tan\frac{x}{\sqrt[]{3}}}=2\\ \Rightarrow \frac{mx}{\frac{x}{\sqrt[]{3}}}\times \left(1\right)=2\Rightarrow \sqrt[]{3}m=2\Rightarrow m=\frac{2}{\sqrt[]{3}}=\frac{2\sqrt[]{3}}{3}.\\ Hence,thevalueofthefilleris\frac{2\sqrt[]{3}}{3}.\end{array}$

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

Sol:

$\begin{array}{l}Givenf\left(x\right)=\underset{x\to \pi }{lim}\frac{-tan\left(\pi -x\right)}{x-\pi }\\ =\underset{\pi -x\to 0}{lim}\frac{-tan\left(\pi -x\right)}{-\left(\pi -x\right)}=1\\ Hence,thevalueofthefilleris1.\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=1-x+x^2-x^3+\dots -x^{99}+x^{100}\\ \therefore f\text{'}\left(x\right)=-1+2x-3x^2+\dots -99x^{98}+100x^{99}\\ So,f\text{'}\left(1\right)=-1+2-3+\dots -99+100\\ =\left(-1-3-5\dots -99\right)+\left(2+4+6+\dots +100\right)\\ =\frac{50}{2}\left[2\times -1+\left(50-1\right)\left(-2\right)\right]+\frac{50}{2}\left[2\times 2+\left(50-1\right)\left(2\right)\right]\\ =25\left[-2-98\right]+25\left[4+98\right]=25\times -100+25\times 102\\ =25\left[-100+102\right]=25\times 2=50.\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=x^{100}+x^{99}+\dots +x+1\\ \therefore f\text{'}\left(x\right)=100x^{99}+99x^{98}+\dots +1\\ So,f\text{'}\left(1\right)=100+99+98+\dots +1\\ =\frac{100}{2}\left[2\times 100+\left(100-1\right)\left(-1\right)\right]=50\left[200-99\right]=50\times 101=5050.\\ Hence,thecorrectoptionis\left(a\right).\end{array}$