# $\lim\limits_{x\to a}\large\frac{\log(x-a)}{\log(e^x-e^a)}$ is

$(a)\;1\qquad(b)\;-1\qquad(c)\;0\qquad(d)\;None\;of\;these$

Using L Hospitals rule $\big(\large\frac{\infty}{\infty}\big)$
$\Rightarrow \lim\limits_{x\to a}\large\frac{1}{x-a}.\frac{e^x-e^a}{e^x}\big(\large\frac{\infty}{\infty}\big)$
$\Rightarrow \lim\limits_{x\to a}\large\frac{e^x}{e^x+(x-a)e^{\large x}}$
$\Rightarrow \large\frac{e^a}{e^a}$
$\Rightarrow 1$
Hence (a) is the correct answer.