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