## Calculus (3rd Edition)

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.