<s>[INST] Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$ 
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
  without use Taylor serie and L'Hôpital.


$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$
now what ?
 [/INST] The hint.
Prove that for all $x>0$ we have:
$$-\frac{1}{6}<\frac{\sin{x}-x}{x^3}<-\frac{1}{6}+\frac{x^2}{120}$$ and from this we obtain:
$$\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}.$$
Let $f(x)=\sin{x}-x+\frac{1}{6}x^3$, where $x>0$.
Thus, $f'''(x)=-\cos{x}+1\geq0$, which gives 
$f''(x)>f''(0)=0$, $f'(x)>f'(0)=0$ and $f(x)>f(0)=0$,
which gives a proof of the left inequality.
By the same way we can prove the right inequality.
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