Câu 5
Ta có $\dfrac{\tan x -\sin x}{x^3}=\dfrac{\sin x(1-\cos x)}{x^3 . \cos x}=\dfrac{2\sin x \sin^2 \dfrac{x}{2}}{x^3 \cos x}=\dfrac{\sin x}{x}. \dfrac{1}{\cos x} .\dfrac{\sin^2 \dfrac{x}{2}}{2.\dfrac{x^2}{4}}=A$
Vậy $\lim \limits_{x\to 0} A = \dfrac{1}{2}$