Question

Evaluate ${lim}_{x\to 0}\frac{\log x+\log \frac{1+x}{x}}{x}$

Answer 1

We have,

${lim}_{x\to 0}\frac{logx+log\frac{1+x}{x}}{x}$

${lim}_{x\to 0}\frac{logx+log(1+x)-logx}{x}$

${lim}_{x\to 0}\frac{log(1+x)}{x}$

Apply L-hospital rule

${lim}_{x\to 0}\frac{\frac{1}{(1+x)}\times 1}{1}$

${lim}_{x\to 0}\frac{1}{(1+x)}$

$=\frac{1}{(1+0)}$

$=1$

Hence, this is the answer.