Find $$\lim\limits_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}$$
So my first thought was to simplify the numerator: $$\tan x-\sin x=\dfrac{\sin x}{\cos x}-\sin{x}=\dfrac{\sin x-\sin x\cos x}{\cos x}=\dfrac{\sin x(1-\cos x)}{\cos x}$$ Now let's plug it back: $$\lim_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}=\lim_{x\to0}{\dfrac{x^3\sin x(1-\cos x)}{\cos x}}$$ I don't see how to use the limit $$\lim_{x\to0}{\dfrac{\sin x}{x}}=1$$
Best Answer
Hint: $$\lim_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3\cos x}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos^2 x)}{x^3\cos x(1+\cos x)}}=\lim_{x\to0}{\dfrac{\sin^3 x}{x^3\cos x(1+\cos x)}}$$