Answer
$$
\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}}=\frac{1}{2}
$$
Work Step by Step
$$\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}}
$$
Since
$$
\lim _{x \rightarrow 0}( e^{x}-1-x)=1-1-0=0
$$
and
$$
\lim _{x \rightarrow 0}(x^{2})=0
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} &\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{e^{x}-1}{2 x} \\
&\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{e^{x}}{2}\\
&=\frac{1}{2}
\end{aligned}
$$
