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 ?

$$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\=  \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} -  \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$
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 ?

