Answer
$$-\frac{7}{3}$$
Work Step by Step
We have
$$
\lim _{x \rightarrow -\infty} \frac{7x^2+4x}{9-3x^2}=-\frac{\infty}{\infty}.
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow -\infty} \frac{7x^2+4x}{9-3x^2}=\lim _{x \rightarrow -\infty} \frac{14x+4}{-6x}=-\frac{\infty}{\infty} .
$$
Applying L’Hôpital’s Rule, we get
$$
\lim _{x \rightarrow -\infty} \frac{14x+4}{-6x}= \lim _{x \rightarrow -\infty} \frac{14}{-6}= -\frac{7}{3}.
$$
