lim _{x rightarrow 0} left[ frac{1}{x} - frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given limit: $L = \lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = \lim _{x \rightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Given limit: $L = \lim _{x ightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} ight] = \lim _{x ightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x - 1)} ight]$ This is in the $\frac{0}{0}$ indeterminate form. Applying $L'	ext{Hopital's rule}$: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(xe^x - x)} = \lim _{x ightarrow 0} \frac{e^x - 1}{e^x + xe^x - 1}$ This is still in the $\frac{0}{0}$ form. Applying $L'	ext{Hopital's rule}$ again: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(e^x + xe^x - 1)} = \lim _{x ightarrow 0} \frac{e^x}{e^x + e^x + xe^x} = \lim _{x ightarrow 0} \frac{e^x}{2e^x + xe^x}$ Substituting $x = 0$: $= \frac{e^0}{2e^0 + 0 \cdot e^0} = \frac{1}{2 + 0} = \frac{1}{2}$

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $

Similar Questions

$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = $

$\lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module