Answer
$$\frac{3}{4}$$
Work Step by Step
We have
$$
\lim _{x \rightarrow \infty} \frac{3x^3+4x^2}{4x^3-7}=\frac{\infty}{\infty}.
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty} \frac{9x^2+8x}{12x^2}=\lim _{x \rightarrow \infty} \frac{9+8/x}{12}=\frac{9}{12}=\frac{3}{4}.
$$
