<s>[INST] Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
 [/INST] $$\frac{\tan ^3 x - \sin ^3 x}{x^5}=\tan^3x\dfrac{1 - \cos ^3 x}{x^5}=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2}\right)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{1-\cos x}{x^2}\right)(1+\cos x+\cos^2x)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{\sin x}{x}\right)^2\left(\dfrac{1}{1+\cos x}\right)(1+\cos x+\cos^2x)$$
Now use  the fundamental limits:
$$\lim_{x\to 0}\frac{\sin x }{x}=1=\lim_{x\to 0}\frac{\tan x }{x}$$
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