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?

$$\dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\ldots=\frac{\sin^3x\left(1-\cos^3x\right)}{x^5\cos^3x}=\tan^3x \cdot\frac{1-\cos^3x}{x^5}$$ $$\sim x^3 \cdot\frac{1-\cos^3x}{x^5}=\frac{1-\cos^3x}{x^2}\sim \frac{1-\left(1-x^2/2+o(x^2)\right)^3}{x^2}$$ $$=\frac{3x^2/2+o(x^2)}{x^2}=\frac{3}{2}+\frac{o(x^2)}{x^2}\underbrace{\to}_{\text{ as }x\to 0}\frac{3}{2}+0=\frac{3}{2}$$
