Answer
The limit does not exist.
Work Step by Step
We have
$$
\lim _{x \rightarrow 0}\frac{e^{2x}-1-x}{ x^2}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0}\frac{e^{2x}-1-x}{ x^2}=\lim _{x \rightarrow 0}\frac{2e^{2x}-1}{ 2x}$$
by checking the one-sidded limits we have
$$
\lim _{x \rightarrow 0^+}\frac{2e^{2x}-1}{ 2x}=\infty \neq \lim _{x \rightarrow 0^-}\frac{2e^{2x}-1}{ 2x}=-\infty $$
hence the limit does not exist.
