Question

$\underset{x\to 1}{lim}\frac{logx}{x-1}$

Answer 1

We have, $\underset{x\to 1}{lim}\frac{logx}{x-1}$

At x=0,the value of the expression is the form $\frac{0}{0}$

applying L'Hospital Rule

$\underset{x\to 1}{lim}\frac{\frac{d}{dx}logx}{\frac{d}{dx}(x-1)}$

$\underset{x\to 1}{lim}\frac{\frac{1}{x}}{1}$ $\{\frac{d}{dx}logx=\frac{1}{x}\}$

Putting x=1

=$\frac{1}{1}=1$