Answer
Indeterminate form is $\dfrac{\infty}{\infty}$ and$$\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1 } = \frac{3 }{5 }$$
Work Step by Step
Given $$\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1}$$
using the method of replacement
\begin{align*}
\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1 }&=\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{3 x^{2}+2 x+4}{5 x^{2}+x+1 }&=\lim _{x \rightarrow \infty} \frac{6 x+2 }{10x+1 } \ \ \text{apply L'Hôpital's Rule}\\
&=\lim _{x \rightarrow \infty} \frac{6 }{10 }\\
&= \frac{6 }{10 }= \frac{3 }{5 }
\end{align*}
