<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] HINT:
$\lim_{x\to0}\left(\left(\dfrac{\tan x}x\right)^3\cdot\dfrac1{x^2}-\left(\dfrac{\sin x}x\right)^3\cdot\dfrac1{x^2}\right)$ is of the form $\infty-\infty$
See List of indeterminate forms
Use 
$$\dfrac1{\cos^3x}\cdot\left(\dfrac{\sin x}x\right)^3\cdot\dfrac{1-\cos x}{x^2}\cdot(1+\cos x+\cos^2x)$$
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