Differentiation (Part-38) | S N De

 

$\,14.\,$ Evaluate :

$\,(i)\,\lim_{x  \to 0}\frac{\sin x}{ \log_e(1+x)^{\frac 12}}\\=\lim_{x  \to 0}\frac{\sin x}{\frac 12 \log_e(1+x)}\\=2 \times \frac{\lim_{x  \to 0}\frac{\sin x}{x}}{\lim_{x  \to 0}\frac{\log_e(1+x)}{x}}\\=2 \times \frac{1}{1}\\=2\,\,\text{(proved)}$

$\,(ii)\,\lim_{x  \to 0}\frac{e^{x^2}-1}{\sin^2x}\\=\lim_{x  \to 0}\frac{\frac{e^{x^2}-1}{x^2}}{\frac{\sin^2x}{x^2}}\\=\lim_{z  \to 0}\frac{e^z-1}{z}\times \frac{1}{\left(\lim_{x  \to 0}\frac{\sin x}{x}\right)^2}\\~~[\text{let}~z=x^2\,\,\text{so that as}~~ x \to 0,~ z\to 0]\\=1 \times  \frac{1}{1^2}\\=1\,\,\text{(ans.)}$

$\,(iii)\lim_{x  \to 0}\,\frac{(e^x-1)\log(1+x)}{\sin x}\\=\lim_{x  \to 0}\left(\frac{e^x-1}{x}\right) \times \frac{1}{\lim_{x  \to 0}\frac{\sin x}{x}} \\~~~~~~~~\times \lim_{x  \to 0}\log(1+x)\\=1 \times \frac 11\times \log(1+0)\\=1 \times \log 1\\=0\,\,\text{(ans.)}$

$\,(iv)\,\lim_{x  \to 4}\frac{x^{7/2}-4^{7/2}}{\log_e(x-3)}\\=\lim_{x  \to 4}\frac{x^{7/2}-4^{7/2}}{x-4} \times \lim_{x  \to 4}\frac{x-4}{\log_e(x-3)}\\=\left(\frac 72 \times 4^{7/2-1}\right) \times  \frac{1}{\lim_{x  \to 4}\frac{\log_e[1+(x-4)]}{x-4}}~~[*]\\=\left(\frac 72 \times 2^{2 \times \frac 52}\right)\times \frac{1}{\lim_{z  \to 0}\frac{\log_e(1+z)}{z}}~~[**]\\=(\frac 72 \times 2^5) \times \frac 11\\= 7\times 2^4\\=112\,\,\text{(ans.)}$

Note[*] $\,\lim_{x  \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$

Note[**]: $\,\text{let}\,\, z=x-4\,\,\text{so that as}~~\,x \to 4,\, z \to 0.$

$\,(v)\,\lim_{x  \to 0}\frac{\log(1+\alpha x)}{\sin\beta x}\\=\lim_{x  \to 0}\frac{\frac{\log(1+\alpha x)}{\alpha x} \times\alpha x}{\frac{\sin\beta x}{\beta x} \times \beta x}\\=\frac{\alpha}{\beta} \times \lim_{(\alpha x)  \to 0} \frac{\log(1+\alpha x)}{\alpha x} \times \frac{1}{\lim_{(\beta x)  \to 0}\frac{\sin\beta x}{\beta x}}\\=\frac{\alpha}{\beta} \times 1 \times \frac 11 \\=\frac{\alpha}{\beta}\,\,\text{(ans.)}$

$\,(vi)\,\lim_{x  \to 0}\frac{e^{3x}-1}{\log(1+5x)}\\=\lim_{x  \to 0}\frac{\frac{e^{3x}-1}{3x} \times 3}{\frac{\log(1+5x)}{5x}\times 5}\\=\frac 35 \times \lim_{(3x)  \to 0} \frac{e^{3x}-1}{3x} \times \frac{1}{\lim_{(5x)  \to 0}\frac{\log(1+5x)}{5x}}\,[*]\\=\frac 35 \times 1 \times \frac 11 \\=\frac 35\,\,\text{(ans.)}$

Note[*] : $\,\lim_{z  \to 0}\frac{e^z-1}{z}=1,\\\lim_{z  \to 0}\frac{\log(1+z)}{z}=1.$

$\,(vii)\,\lim_{x \to  0}\frac{2^{3x}-1}{3^{2x}-1}\\=\lim_{x \to  0}\frac{\frac{2^{3x}-1}{3x}\times 3}{\frac{3^{2x}-1}{2x} \times 2}\\=\frac 32 \times  \lim_{z  \to 0}\frac{2^z-1}{z} \times  \frac{1}{\lim_{p \to 0}\frac{3^p-1}{p}}~~[*]\\=\frac 32 \times  \log 2 \times \frac{1}{\log 3}\\=\frac{3\log 2}{2 \log 3}\\=\frac{\log 2^3}{\log 3^2}\\=\frac{\log 8}{\log 9}\,\,\text{(ans.)}$

Note [*] : Here, let $\,z=3x,\,p=2x\,\,\text{and as}~~\,x \to 0,z \to 0,\,p \to 0.$

$\,15.\,$ Find the limit (if exists): $\,\lim_{x \to  0}\frac{|x|}{x}$

Sol. We know, $\,|x|=x,\,~~~x \geq 0 \\~~~~~=-x,\,x <0$

L.H.L$=\lim_{x \to  0-}\frac{|x|}{x}=\lim_{x \to  0-}\frac{-x}{x}=-1$

R.H.L. $=\lim_{x \to  0+}\frac{|x|}{x}=\lim_{x \to  0+}\frac{x}{x}=1$

Hence, L.H.L $\,\neq\,$ R.H.L.

So,  $\,\lim_{x \to  0}\frac{|x|}{x}\,\,$ does not exist.

$\,16.\,$ Find the limit (when it exists) : $\,\lim_{x \to  0}\frac{\sin|x|}{x}.$

Sol. We know, $\,|x|=x,\,~~~x \geq 0 \\~~~~~=-x,\,x <0$

So, R.H.L $=\lim_{x \to  0+}\frac{\sin|x|}{x}=\lim_{x \to  0+}\frac{\sin x}{x}=1$

L.H.L. $=\lim_{x \to  0-}\frac{\sin|x|}{x}\\=\lim_{x \to  0-}\frac{\sin(-x)}{x}=- \lim_{x \to  0-}\frac{\sin x}{x}=-1$

$\,\therefore \,\, \text{R.H.L.} \neq \text{L.H.L.}$

So, $\,\lim_{x \to  0}\frac{\sin|x|}{x}.$ does not exist.

Note : Here, R.H.L. denotes Right Hand Limit and L.H.L. denotes Left Hand Limit.

$\,17.\,$ Find the limit (if exists): $\,\lim_{x \to  0}\frac{x[x]}{\sin|x|}\,;$ (where $\,[x]\,$ denotes the largest integer not greater than $\,x$)

Sol. We know, $\,|x|=x,\,~~~x \geq 0 \\~~~~~=-x,\,x <0$

Also, $\,[x]=-1,\,~ -1 \leq x <0\\~~~~~=0,\,~~~~~~~~~0 \le x <1$

So, R.H.L.$=\lim_{x \to  0+}\frac{x[x]}{\sin|x|}\\=\lim_{x \to  0+}\frac{x \times 0}{\sin x}\\=\lim_{x \to  0+}\frac{0}{\sin x}\\=0$

and L.H.L.$=\lim_{x \to  0-}\frac{x[x]}{\sin|x|}\\=\lim_{x \to  0-}\frac{x \times (-1)}{\sin (-x)}\\=\lim_{x \to  0-}\frac{x}{\sin x}\\=\frac{1}{\lim_{x \to  0-}\frac{\sin x}{x}}\\=\frac 11 \\=1$

Hence, L.H.L $\,\neq\,$ R.H.L.

So,   $\,\lim_{x \to  0}\frac{x[x]}{\sin|x|}\,$  does not exist.

To download full PDF on LIMIT , click here .

$\,18.\,$ Find the limit (if exists) : $\,\lim_{x  \to 0}\frac{\sin \{x\}}{\{x\}},\,$ (where $\,\{x\}\,$ denotes the fractional part of $\,x\,$)

Sol.  $\,[x]=-1,\,~ -1 \leq x <0\\~~~~~=0,\,~~~~~~~~~0 \le x <1$

$\,\therefore \,\lim_{x \to  0+}\frac{\sin \{x\}}{\{x\}}\\=\lim_{x \to  0+}\frac{\sin(x-[x])}{x-[x]}\\=\lim_{x \to  0+}\frac{\sin(x-0)}{x-0}\\=\lim_{x \to  0+}\frac{\sin x}{x}\\=1$ 

$\,\text{Also,}~~ \,\lim_{x \to  0-}\frac{\sin \{x\}}{\{x\}}\\=\lim_{x \to  0-}\frac{\sin(x-[x])}{x-[x]}\\=\lim_{x \to  0-}\frac{\sin(x+1)}{x+1}\\=\lim_{x \to  0-}\frac{\sin (0+1)}{0+1}\\=\sin 1$ 

$\,\therefore \,\lim_{x \to  0+}\frac{\sin \{x\}}{\{x\}}\neq \lim_{x \to  0-}\frac{\sin \{x\}}{\{x\}}$

So,   $\,\lim_{x  \to 0}\frac{\sin \{x\}}{\{x\}}\,$ does not exist.

$\,19.\,$ Evaluate : $\, \lim_{x  \to 2-0}\{x+(x-[x]^2)\}\,\,$, (where $\,[x]\,$ denotes the largest integer not greater than $\,x.$)

Sol. If $~\,x \to 2-0,\,~~1<x<2 \\ \therefore [x]=1.$

Now, $\,~~~\lim_{x  \to 2-0} \{x+(x-[x]^2)\}\\=\lim_{x  \to 2-0}\{x+(x-1^2)\}\\=\lim_{x  \to 2-0}\{2x-1\}\\=2 \times 2-1\\=3.$