Toán Tính: $lim_{x->0}\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}$ 07/07/2021 By Josie Tính: $lim_{x->0}\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}$
Lời giải: Ta có: $\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}$$=\frac{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}+x^5\epsilon(x))-\frac{1}{2}.(x-\frac{x^3}{3!}+x^4\epsilon_2(x))^2}{x^4}$$=\frac{\frac{1}{8}x^4+x^5\epsilon(x)}{x^4}$~$\frac{\frac{1}{8}x^4}{x^4}→\frac{1}{8}$ khi $x→0$Vậy $\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}=\frac{1}{8}$ Trả lời
Lời giải:
Ta có:
$\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}$
$=\frac{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}+x^5\epsilon(x))-\frac{1}{2}.(x-\frac{x^3}{3!}+x^4\epsilon_2(x))^2}{x^4}$
$=\frac{\frac{1}{8}x^4+x^5\epsilon(x)}{x^4}$~$\frac{\frac{1}{8}x^4}{x^4}→\frac{1}{8}$ khi $x→0$
Vậy $\frac{1-cosx-\frac{1}{2}sin^2x}{x^4}=\frac{1}{8}$
Đáp án:nt