Question

Solve $\underset{x\to 0}{lim}\{\frac{{\log }_c\left(1+x\right)}{x^2}+\frac{x-1}{x}\}=$

Answer 1

We have,

${lim}_{x\to 0}\{\frac{{\log }_e(1+x)}{x^2}+\frac{x-1}{x}\}$

${lim}_{x\to 0}\left\{\frac{{\log }_e(1+x)}{x^2}\right\}+{lim}_{x\to 0}\left\{\frac{x-1}{x}\right\}$

By using L-hospital rule,

${lim}_{x\to 0}\left\{\frac{1}{2x(1+x)}\right\}+{lim}_{x\to 0}\left\{\frac{1-0}{1}\right\}$

${lim}_{x\to 0}\left\{\frac{1}{2x+2x^2}\right\}+{lim}_{x\to 0}\left\{1\right\}$

$\frac{1}{0}+1$

$=\infty$

Hence, the value is $\infty$.