Answer
Indeterminate form is $\dfrac{0}{0}$ and$$ \lim _{x \rightarrow \infty} \frac{4 x^{3}}{5 x^{3}+6} =\frac{4}{5}$$
Work Step by Step
Given $$ \lim _{x \rightarrow \infty} \frac{4 x^{3}}{5 x^{3}+6} $$
using the method of replacement
\begin{align*}
\lim _{x \rightarrow \infty} \frac{4 x^{3}}{5 x^{3}+6} &=\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^{3}}{5 x^{3}+6} &= \lim _{x \rightarrow \infty} \frac{12x^{2}}{15 x^{2} } \\
=&\lim _{x \rightarrow \infty} \frac{12 }{15 }\\
&=\frac{4}{5}
\end{align*}
