Tính: $lim_{x->0}(\frac{1}{sin^2x}-\frac{1}{x^2})$ 05/07/2021 Bởi Clara Tính: $lim_{x->0}(\frac{1}{sin^2x}-\frac{1}{x^2})$
Lời giải: Ta có:$\frac{1}{sin^2x}-\frac{1}{x^2}=\frac{x^2-sin^2x}{x^2sin^2x}=\frac{x^2-(x-\frac{x^3}{6}+x^4\epsilon(x))^2}{x^2sin^2x}$~$\frac{\frac{1}{3}x^4}{x^4}->\frac{1}{3}$ khi $x->0$Vậy $lim_{x->0}(\frac{1}{sin^2x}-\frac{1}{x^2})=\frac{1}{3}$ Bình luận
Đáp án:???
Lời giải:
Ta có:
$\frac{1}{sin^2x}-\frac{1}{x^2}=\frac{x^2-sin^2x}{x^2sin^2x}=\frac{x^2-(x-\frac{x^3}{6}+x^4\epsilon(x))^2}{x^2sin^2x}$~$\frac{\frac{1}{3}x^4}{x^4}->\frac{1}{3}$ khi $x->0$
Vậy $lim_{x->0}(\frac{1}{sin^2x}-\frac{1}{x^2})=\frac{1}{3}$