How to solve $\lim_{x\to0^+} \frac{\sqrt{\sin x}}{\sin\sqrt{x}}$ without the use of L'Hospital's Rule I have tried manipulating the expression to get rid of the indeterminate form. I tried to use intervals and the sandwich theorem. 
 A: Since $(\sin x)/x\to1$ as $x\to0$, we have that
$$
\frac{\sqrt{\sin x}}{\sin\sqrt x}=\frac{\sqrt{x}\sqrt{(\sin x)/x}}{\sqrt{x}(\sin\sqrt x)/\sqrt{x}}=\frac{\sqrt{(\sin x)/x}}{(\sin\sqrt x)/\sqrt{x}}\to1
$$
as $x\to0$.
A: hint
$$\frac{\sqrt{\sin(x)}}{\sin(\sqrt x)}=\sqrt{\frac{\sin(x)}{x}}\cdot \frac{\sqrt x}{\sin(\sqrt{x})}$$
A: Since $f(x)=\sqrt{x}$ is a continuous function for $x\rightarrow0^+$, we obtain:
$$\lim_{x\rightarrow0^+}\frac{\sqrt{\sin{x}}}{\sin{\sqrt{x}}}=\lim_{x\rightarrow0^+}\left(\sqrt{\frac{\sin{x}}{x}}\cdot\frac{\sqrt{x}}{\sin{\sqrt{x}}}\right)=\sqrt{\lim_{x\rightarrow0^+}\frac{\sin{x}}{x}}\cdot\lim_{x\rightarrow0^+}\frac{\sqrt{x}}{\sin{\sqrt{x}}}=\sqrt1\cdot1=1.$$
A: It's
$$\sqrt{\frac{\sin x}x}\left(\frac{\sin\sqrt x}{\sqrt x}\right)^{-1}$$
isn't it?
A: i would write
$$\sqrt{\frac{\sin(x)}{x}}\cdot \frac{1}{\frac{\sin(\sqrt{x})}{\sqrt{x}}}$$
A: An alternative method would be to use the following Taylor expansions:
$$\sin(x)=x+O(x^3)$$
$$\sin(\sqrt{x})=\sqrt{x}+O(\sqrt{x}^3).$$
You will obtain:
$$\lim_{x\to 0^+} \frac{\sqrt{\sin x}}{\sin\sqrt{x}}=\lim_{x\to 0^+} \frac{\sqrt{x+O(x^3)}}{\sqrt{x}+O(\sqrt{x}^3)}=\lim_{x\to 0^+} \frac{\sqrt{1+O(x^{2})}}{1+O(x)}=1$$
A: $$\lim_{x\to 0^+} \frac{\sqrt{\sin x}}{\sin\sqrt{x}}$$
First you must know the fundamental limits
$$\lim_{t\to 0}\frac{\sin t}{t}=\lim_{t\to 0}\frac{t}{\sin t}=1$$
Then how to use it in this limit
I see a $\sqrt{\sin x}$ in the numerator so I need a $\sqrt{x}$ in the numerator. To compensate I multiply denominator by $\sqrt x$ and I get
$$\lim_{x\to 0^+} \frac{\sqrt{\sin x}}{\sin\sqrt{x}}=\lim_{x\to 0^+}\frac{\color{red}{\sqrt x}\sqrt{\sin x}}{\color{red}{\sqrt x}\sin \sqrt x}=\lim_{x\to 0^+}\frac{\sqrt x}{\sin \sqrt x}\cdot \lim_{x\to 0^+}\frac{\sqrt {\sin x}}{\sqrt x}=1\cdot  \lim_{x\to 0^+}\sqrt{\frac{ {\sin x}}{x}}=1\cdot 1$$ 
Hope this is useful
