Answer
Indeterminate form is $\dfrac{\infty}{\infty}$ and $$\lim _{x \rightarrow \infty} \frac{4 x^{2}+7}{2 x^{3}+3}=0$$
Work Step by Step
Given $$\lim _{x \rightarrow \infty} \frac{4 x^{2}+7}{2 x^{3}+3}$$
using the method of replacement
\begin{align*}
\lim _{x \rightarrow \infty} \frac{4 x^{2}+7}{2 x^{3}+3}&=\frac{\infty}{\infty}
\end{align*}
This is an indeterminate form, then applying L'Hôpital's Rule, we
get
\begin{align*}
\lim _{x \rightarrow \infty} \frac{4 x^{2}+7}{2 x^{3}+3}&=\lim _{x \rightarrow \infty} \frac{8 x}{6x^{2}} \ \ \text{apply L'Hôpital's Rule}\\
&=\lim _{x \rightarrow \infty} \frac{8 }{12x }\\
&=\lim _{x \rightarrow \infty} \frac{8}{\infty }\\
&=0\end{align*}
