Question

$\underset{x\to 1}{lim}[\frac{1}{\ln x}-\frac{1}{(x-1)}]$

• A. Does not exist
• B. $1$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{2}$

$\underset{x\to 1}{lim}(\frac{1}{\log x}-\frac{1}{(x-1)})$
$\Rightarrow \underset{x\to 1}{lim}\left(\frac{x-1-\log x}{(x-1)\log x}\right)$
Using L'Hospital's Rule, we get
$\Rightarrow \underset{x\to 1}{lim}\left(\frac{1-0-\frac{1}{x}}{\log x+\frac{x-1}{x}}\right)$
Using L'Hospital's Rule again, we get
$\Rightarrow \underset{x\to 1}{lim}\left\{\frac{\frac{1}{x^2}}{\frac{1}{x}+\frac{1}{x^2}}\right\}=\frac{1}{2}$