\[\begin{array}{l}
limxti?nt?i1{x^1}0 - 1/{x^2} - 1limxti?nt?i4{x^2} - 16/{x^2} + x - 20\\
{\lim _{x \to 1}}\frac{{{x^{10}} - 1}}{{{x^2} - 1}}\\
= {\lim _{x \to 1}}\frac{{\left( {{x^5} - 1} \right)\left( {{x^5} + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= {\lim _{x \to 1}}\frac{{\left( {x - 1} \right)\left( {{x^4} + {x^3} + {x^2} + x + 1} \right)\left( {{x^5} + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= {\lim _{x \to 1}}\frac{{\left( {{x^4} + {x^3} + {x^2} + x + 1} \right)\left( {{x^5} + 1} \right)}}{{x + 1}}\\
= \frac{{\left( {{1^4} + {1^3} + {1^2} + 1 + 1} \right)\left( {{1^5} + 1} \right)}}{{1 + 1}}\\
= \frac{{5.2}}{2}\\
= 5\\
{\lim _{x \to 4}}\frac{{{x^2} - 16}}{{{x^2} + x - 20}}\\
= {\lim _{x \to 4}}\frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 5} \right)}}\\
= {\lim _{x \to 4}}\frac{{x + 4}}{{x + 5}}\\
= \frac{{4 + 4}}{{4 + 5}}\\
= \frac{8}{9}
\end{array}\]