<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] note that 
$$
\begin{split}
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x^2}&\overset{t=\sqrt{x}}{=}
\lim_{t\to0^{+}}
\frac{\sin(t^2)-\sin^2t}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\ldots)-(t^2-\frac{t^4}{3}+\frac{2t^{6}}{45}+\frac{t^8}{315}-\ldots}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{\frac{t^4}{3}-\frac{19t^6}{90}+\frac{t^8}{315}+\ldots)}{t^4}\\
&=\frac{1}{3}
\end{split}
$$
Hence
$$
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin\sqrt{x})}=
\lim_{x\to0^{+}}
\frac{\frac{\sin x-\sin^2\sqrt{x}}{x^2}}{\sqrt{\frac{\sin x}{x}}+\frac{\sin\sqrt{x}}{\sqrt{x}}}
=\frac{\frac{1}{3}}{\sqrt{1}+1}=\frac{1}{6}
$$
</s>