Example 2. Find $\lim _{x \rightarrow 0} \frac{\sin x e^{2 x}-x}{5 x^{2}+x^{3}}$.
Solution. In this case, the L'Hôpital-Bernoulli rule needs to be applied twice, as the ratio of the first derivatives again represents an indeterminate form of $\frac{0}{0}$.

Indeed,

$$
\lim _{x \rightarrow 0} \frac{\sin x e^{2 x}-x}{5 x^{2}+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(\sin x e^{2 x}-x\right)^{\prime}}{\left(5 x^{2}+x^{3}\right)^{\prime}}=\lim _{x \rightarrow 0} \frac{\cos x e^{2 x}+2 \sin x e^{2 x}-1}{10 x+3 x^{2}}
$$

As $x \rightarrow 0$, we again obtain an indeterminate form of $\frac{0}{0}$. Applying the L'Hôpital-Bernoulli rule once more, we find

$$
\lim _{x \rightarrow 0} \frac{\cos x e^{2 x}+2 \sin x e^{2 x}-1}{10 x+3 x^{2}}=\lim _{x \rightarrow 0} \frac{\left(\cos x e^{2 x}+2 \sin x e^{2 x}-1\right)^{\prime}}{\left(10 x+3 x^{2}\right)^{\prime}}=
$$

$$
=\lim _{x \rightarrow 0} \frac{-\sin x e^{2 x}+4 \cos x e^{2 x}+4 \sin x e^{2 x}}{10+6 x}=\frac{4}{10}=\frac{2}{5}
$$