<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] $$\begin{align}
\lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}
&=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\
&=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\infty}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\left(1+\frac12x^2+\frac5{24}x^4+\cdots\right)^{3/2}-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\left(1+\frac34x^2+\cdots\right)-1}{x^2}\\
&=\frac34
\end{align}$$
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