Divide everything by the highes power of x, so \(\displaystyle{x}^{{3}}\)

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}=\lim_{{{x}\to\infty}}{\frac{{{\frac{{{1}}}{{{x}^{{3}}}}}-{\frac{{{1}}}{{{x}}}}}}{{{1}-{\frac{{{1}}}{{{x}^{{2}}}}}+{\frac{{{1}}}{{{x}^{{3}}}}}}}}\)

Then as \(\displaystyle{x}\to\infty\) terms \(\displaystyle{\frac{{{1}}}{{{x}^{{3}}}}},{\frac{{{1}}}{{{x}^{{2}}}}},{\frac{{{1}}}{{{x}}}}\) approach 0.

\(\displaystyle{\frac{{{0}-{0}}}{{{1}-{0}+{0}}}}={\frac{{{0}}}{{{1}}}}={0}\)

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}={0}\)

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}=\lim_{{{x}\to\infty}}{\frac{{{\frac{{{1}}}{{{x}^{{3}}}}}-{\frac{{{1}}}{{{x}}}}}}{{{1}-{\frac{{{1}}}{{{x}^{{2}}}}}+{\frac{{{1}}}{{{x}^{{3}}}}}}}}\)

Then as \(\displaystyle{x}\to\infty\) terms \(\displaystyle{\frac{{{1}}}{{{x}^{{3}}}}},{\frac{{{1}}}{{{x}^{{2}}}}},{\frac{{{1}}}{{{x}}}}\) approach 0.

\(\displaystyle{\frac{{{0}-{0}}}{{{1}-{0}+{0}}}}={\frac{{{0}}}{{{1}}}}={0}\)

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}={0}\)