<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] $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}$$
with $\cos x\approx 1-\frac12x^2$ as $x\to0$
$$\lim_{x\to 0} \dfrac{\sin ^3 x}{x^3}\times\dfrac{1-(1-\frac12x^2)^3}{x^2\times1}=\lim_{x\to 0}\frac32-\frac34x^2+\frac18x^4=\frac32$$
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