calculus - Calculating $lim_{x rightarrow 0} frac{tan x - sin x}{x^3}$.

I have a difficulty in calculating this limit:

$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$

I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$

Then I got stucked, could anyone help me in solving it?

Answer

For $x\ne0,$

$${\tan x-\sin x\over x^3}=\left({\sin x\over x}\right)^3\dfrac1{\cos x \,(1+\cos x)}$$

Now as $x\to0,x\ne0$