<s>[INST] $\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.
 [/INST] You can first remove a few factors
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}\\
=\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt{\tan x}}{x \sqrt{x}}\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}\\
=\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}.$$
Then multiply by the conjugate 
$$=\lim_{x \to 0^+}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})},$$
evaluate the finite factor at denominator
$$=\frac12\lim_{x \to 0^+}\frac{1-\cos x(1-\sin^2 x)}{x^2},$$
use trigonometric identitites
$$=\frac12\lim_{x \to 0^+}\frac{2\sin^2\frac x2+\cos x\sin^2 x}{x^2},$$
and conclude
$$=\left(\frac12\right)^2+\frac12.$$

We used
$$\frac{\tan x}x=\frac{\sin x}x\frac1{\cos x}\to 1,$$
$$\frac{\sin ax}x=a\frac{\sin ax}{ax}\to a.$$
</s>