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 ?
 A: $$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\=  \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} -  \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$
Let $y = \sqrt{x}$
Then $${\rm L^{\prime\prime}} = \lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} -  \sqrt{x})}{x\sqrt{x}} = \lim\limits_{x \to 0^+}\dfrac{(\sin y -  y)}{y^3} = -\dfrac 16$$
For the other limit,
$${\rm L^{\prime}}=\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}} =\lim\limits_{x \to 0^+}\dfrac{(\sin x- x)}{x^3}\dfrac{x\sqrt{x}}{(\sqrt{\sin x} + \sqrt{x})}= \dfrac {-1}6\lim\limits_{x \to 0^+}\dfrac{x}{\left(\sqrt{\dfrac{\sin x}{x}} + 1\right)} = 0$$
Hence our limit is $\mathrm L = \dfrac {1}{6}$.

I used $$\lim_{x \to 0} \dfrac{\sin x - x}{x^3} = \dfrac{-1}6$$
Are all limits solvable without L'Hôpital Rule or Series Expansion ?

A: 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.
A: 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
$$
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin\sqrt{x})}=
\lim_{x\to0^{+}}
\frac{\frac{\sin x-\sin^2\sqrt{x}}{x^2}}{\sqrt{\frac{\sin x}{x}}+\frac{\sin\sqrt{x}}{\sqrt{x}}}
=\frac{\frac{1}{3}}{\sqrt{1}+1}=\frac{1}{6}
$$
