Answer
$$1$$
Work Step by Step
Given $$ \lim _{x\to \infty } \frac{x^2}{\sqrt{x^4+1}} $$
Then
\begin{aligned}
\lim _{x\to \infty } \frac{x^2}{\sqrt{x^4+1}} &=\lim _{x\to \infty } \frac{\frac{x^2}{x^2}}{\sqrt{\frac{x^4}{x^4}+\frac{1}{x^4}}} \\
&= \lim _{x\to \infty } \frac{1}{\sqrt{1+\frac{1}{x^4}}} \\
&=\lim _{x\to \infty } \frac{1}{\sqrt{1+0}}\\
&=1
\end{aligned}