Answer
$$-1$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}+1}}{x+1}&=\lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}+1} / \sqrt{x^{2}}}{(x+1) / \sqrt{x^{2}}}\\
&=\lim _{x \rightarrow-\infty} \frac{\sqrt{\left(x^{2}+1\right) / x^{2}}}{(x+1) /(-x)}\\
&=\lim _{x \rightarrow \infty} \frac{\sqrt{1+1 / x^{2}}}{(-1-1 / x)}\\
&=\frac{\sqrt{1+0}}{(-1-0)}\\
&=-1
\end{align*}