Answer
$$0$$
Work Step by Step
Given $$ \lim _{x\rightarrow \infty}\frac{1-x^2}{x^3-x+1}$$
Then
\begin{aligned}
\lim _{x\rightarrow \infty}\frac{1-x^2}{x^3-x+1} &=
\lim _{x\rightarrow \infty}\frac{\frac{1}{x^3}-\frac{x^2}{x^3}}{\frac{x^3}{x^3}-\frac{x }{x^3}+\frac{1}{x^3}} \\
&= \lim _{x\rightarrow \infty}\frac{0-0}{1-0+0}\\
&=0
\end{aligned}