Answer
$\infty $
Work Step by Step
We have
$$
\lim _{x \rightarrow \infty}\frac{e^{2x}-1-x}{ x^2}=\frac{\infty}{\infty}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty}\frac{e^{2x}-1-x}{ x^2}=\lim _{x \rightarrow \infty}\frac{2e^{2x}-1}{ 2x}=\frac{\infty}{\infty}$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty}\frac{4e^{2x}}{ 2}=\infty.$$
