Answer
$$\infty$$
Work Step by Step
Given
$$\lim _{x \rightarrow \infty}\sqrt{ x^2+1}$$
Then
\begin{aligned}
\lim _{x \rightarrow \infty}\sqrt{ x^2+1}&=\sqrt{\lim _{x \rightarrow \infty} (x^2+1)}\\
&=\sqrt{ \infty}\\
&=\infty
\end{aligned}