The value of mathop {lim }limits_{x to a} fra | vedclass

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

Question

(A) We are given the limit $L = \mathop {\lim }\limits_{x 	o a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$.Since the expression is in the form $\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We are given the limit $L = \mathop {\lim }\limits_{x 	o a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$.Since the expression is in the form $\frac{-\infty}{-\infty}$ as $x 	o a^+$,we apply $L$'$H$ôpital's Rule:$L = \mathop {\lim }\limits_{x 	o a} \frac{\frac{d}{dx}[\log(x-a)]}{\frac{d}{dx}[\log(e^x - e^a)]} = \mathop {\lim }\limits_{x 	o a} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} = \mathop {\lim }\limits_{x 	o a} \frac{e^x - e^a}{(x-a)e^x}$.This is now in the $\frac{0}{0}$ form. Applying $L$'$H$ôpital's Rule again:$L = \mathop {\lim }\limits_{x 	o a} \frac{e^x}{e^x + (x-a)e^x} = \frac{e^a}{e^a + (a-a)e^a} = \frac{e^a}{e^a} = 1$.

The value of $\mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$ is

Similar Questions

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

If $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$,then $\mathop {\lim }\limits_{x \to a} \,\frac{f(a)g(x) - f(x)g(a)}{x - a} = $

If $f$ is a strictly increasing function,then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to

$\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$ is equal to

If $f(5)=7$ and $f'(5)=7$, then $\lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module