Question

Помогите решить задание с пределом

Answer 1

$\displaystyle \lim_{x \to 1}(\frac{x^4-1}{2x^4-x^2-1})\\\\ \lim_{x \to 1}(x^4-1) =0\\ \lim_{x \to 1}(2x^4-x^2-1)=0\\\\ \lim_{x \to 1}(\frac{x^4-1}{2x^4-x^2-1})= \lim_{x \to 1} (\frac{(x^2-1)(x^2+1)}{2x^4+x^2-2x^2-1})= \lim_{x \to 1}(\frac{(x^2-1)(x^2+1)}{x^2(2x^2+1)-(2x^2+1)}= \\ \\ \lim_{x \to 1}(\frac{(x^2-1)(x^2+1)}{(2x^2+1)(x^2-1)})= \lim_{x \to 1}(\frac{x^2+1}{2x^2+1})=\frac{1^2+1}{2*1^2+1}=\frac{2}{3}$