<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] $$\begin{align}\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}&=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \left(\dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\cdot\frac{1+\cos x}{1+\cos x}\right), \quad\text{note that }(1-\cos x)(1+\cos x)=\sin^2x\\
&=\lim_{x\to 0} \dfrac{\sin ^5 x\quad(1+\cos x+\cos^2x)}{x^5\cos ^3 x\quad(1+\cos x)}\\
&=\lim_{x\to 0}\left[\left(\frac{\sin x}{x}\right)^5\cdot \frac{1}{\cos^3x}\cdot\frac{1+\cos x+\cos^2x}{1+\cos x}\right]\\
&=1^5\cdot\frac{1}{1^3}\cdot\frac{1+1+1^2}{1+1}=\frac{3}{2}.
\end{align}$$
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