Example 1. Find $\lim _{x \rightarrow 0} \frac{\sin x e^{x}-5 x}{4 x^{2}+7 x}$.
Solution. When $x \rightarrow 0$, both the numerator and the denominator of the fraction approach zero, resulting in an indeterminate form of $\frac{0}{0}$.

Applying L'Hôpital's rule, we find

$$
\lim _{x \rightarrow 0} \frac{\sin x e^{x}-5 x}{4 x^{2}+7 x}=\lim _{x \rightarrow 0} \frac{\left(\sin x e^{x}-5 x\right)^{\prime}}{\left(4 x^{2}+7 x\right)^{\prime}}=\lim _{x \rightarrow 0} \frac{\cos x e^{x}+\sin x e^{x}-5}{8 x+7}=-\frac{4}{7}
$$