Question

Evaluate :
$\underset{x\to 1}{lim}\frac{\log e^x}{x-1}$

Answer 1

We have,

${lim}_{x\to 1}\frac{\log e^x}{x-1}$

By L-Hospital rule,

${lim}_{x\to 1}\frac{\frac{1}{e^x}\times e^x}{1-0}$

${lim}_{x\to 1}\frac{1}{1}$

${lim}_{x\to 1}1$

$=1$

Hence, the value is $1$.