Question

Solve:
$\underset{x\to 0}{lim}$ $\frac{e^x-1-x}{x^2}$

Answer 1

We have,

${lim}_{x\to 0}\frac{e^x-1-x}{x^2}$

This is the $\frac{0}{0}$ form.

So, apply L-hospital

${lim}_{x\to 0}\frac{e^x-0-1}{2x}$

${lim}_{x\to 0}\frac{e^x-1}{2x}$

This is the $\frac{0}{0}$ form.

So, apply L-hospital

${lim}_{x\to 0}\frac{e^x-0}{2}$

$=\frac{e^0}{2}$

$=\frac{1}{2}$

Hence, this is the answer.