Question

$\underset{x\to 1}{lim}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)=$

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{5}{2}$

Answer 1

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

$\underset{x\to 1}{lim}(\frac{1}{\ln x}-\frac{1}{x-1})={lim}_{x\to 1}\frac{x-1-\ln x}{\ln x(x-1)}$
$=\underset{x\to 1}{lim}\frac{x-1}{-1+x+x\ln x}$ ( Applying L'Hospital rule)
$=\underset{x\to 1}{lim}\frac{1}{2+\ln x}$ (Applying L'Hospital rule)
$=\frac{1}{2}$
Hence, option 'A' is correct.