Question

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

• A. $0$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

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

$L=\underset{x\to 1}{lim}(\frac{1}{\ln x}-\frac{1}{x-1})$ [$\infty -\infty$ form]
Put $x=1+h$

$L=\underset{h\to 0}{lim}(\frac{1}{\ln (1+h)}-\frac{1}{h})$

$=\underset{h\to 0}{lim}\frac{h-\ln (1+h)}{h\ln (1+h)}$

$=\underset{h\to 0}{lim}\frac{h-\ln (1+h)}{h^2\frac{\ln (1+h)}{h}}$

$\Rightarrow L=\underset{h\to 0}{lim}\frac{h-\ln (1+h)}{h^2}\cdots \left(1\right)$ $(\because \underset{h\to 0}{lim}\frac{\ln (1+h)}{h}=1)$

$\Rightarrow L=\underset{h\to 0}{lim}\frac{h-[h-\frac{h^2}{2}+\frac{h^3}{3}-\frac{h^4}{4}+\cdots ]}{h^2}$

$=\underset{h\to 0}{lim}\frac{h^2[\frac{1}{2}-\frac{h}{3}+\frac{h^2}{4}-\cdots ]}{h^2}$
$=\frac{1}{2}$

Alternative Solution :

From equation $\left(1\right)$
by L'Hospital rule,
$L=\underset{h\to 0}{lim}\frac{1-\frac{1}{1+h}}{2h}$
Again differentiation with respect to $h$
$L=\underset{h\to 0}{lim}\frac{0+\frac{1}{(1+h{)}^2}}{2}$
$=\frac{1}{2}$