Question

$\underset{x\to e}{lim}\frac{\ln x-1}{|x-e|}$ is equal to

• A. $1/e$
• B. $-1/e$
• C. $e$
• D. does not exist

Answer 1

The correct option is D does not exist

Let, $L=\underset{x\to e}{lim}\frac{\ln x-1}{|x-e|}$.

Now the right hand limit i.e.

$RHL$ $=\underset{x\to e+}{lim}\frac{\ln x-1}{x-e}$ $\frac{0}{0}$ form

Now L'Hospital's rule we get,

or, $RHL$ $=\underset{x\to e+}{lim}\frac{\frac{1}{x}}{1}=\frac{1}{e}$.......(1).

Again, the right hand limit i.e.

$LHL$ $=\underset{x\to e-}{lim}\frac{\ln x-1}{-x+e}$ $\frac{0}{0}$ form

Now L'Hospital's rule we get,

or, $LHL$ $=\underset{x\to e-}{lim}\frac{\frac{1}{x}}{-1}=-\frac{1}{e}$.......(2).

Since $RHL\ne LHL$, so the given limit doesn't exit.