Calculate $\lim _{x \rightarrow 0} \frac{\int _{0}^{\sin x} \sqrt{\tan x} dx}{\int _{0}^{\tan x} \sqrt{\sin x} dx}$ 
Calculate
  $$
\lim _{x \to 0}\frac{\int_{0}^{\sin\left(x\right)}
\,\sqrt{\,\tan\left(t\right)\,}\,\mathrm{d}t}
{\int_{0}^{\tan\left(x\right)}\,\sqrt{\,\sin\left(t\right)\,}\, \mathrm{d}t}
$$


I have a problem with this task because my first idea was to calculate $\int_{0}^{\sin\left(x\right)}
\,\sqrt{\,\tan\left(t\right)\,}\,\mathrm{d}t$ and $\int_{0}^{\tan\left(x\right)}\,\sqrt{\,\sin\left(t\right)\,}\, \mathrm{d}t$ and then calculate $\lim$. However this calculation are really long and I think that author of this task wanted to use some tricks to don't count this.Have you got some ideas ?.
 A: Hints:
First, it should be 
$$\;\lim_{x\to0}\frac{\int\limits_0^{\sin x}\sqrt{\tan t}\,dt}{\int\limits_0^{\tan x}\sqrt{\sin t}\,dt}$$
Observe that both integrands are continuous functions, and thus their integrals are differentiable functions of the upper limit...
Use now L'Hospital (why can you?), and for example in the numerator get $\;\cos x\sqrt{\tan \sin x}\;$ ...and continue.
A: We'll use Lebnitz and L'Hopital rule. 
$$\lim _{x \rightarrow 0} \frac{\int _{0}^{\sin x} \sqrt{\tan x} dx}{\int _{0}^{\tan x} \sqrt{\sin x} dx}$$
$$= \lim_{x \to 0} \frac{\sqrt{\tan(\sin x)} \cos x}{\sqrt{\sin(\tan x)} \sec^2 x}$$
$$=\lim_{x \to 0} \frac{\sqrt{\tan(\sin x)}}{\sqrt{\sin(\tan x)}}$$
Because $\cos x \to 1$ as $x \to 0$. 
Now just to make life simple we will calculate the limit of the square of the expression and then we'll take the positive square root of the result in the end. So essentially we'll calculate: 
$$\lim_{x \to 0} \frac{\tan(\sin x)}{\sin(\tan x)}$$
Now since the following two identities hold: 
$$\lim_{t \to 0} \frac{\sin t}{t} = 1 \quad (1)$$
$$\lim_{t \to 0} \frac{\tan t}{t} = 1 \quad (2)$$
We will multiply divide by $\sin x$ and $\tan x$.
$$=\lim_{x \to 0} \frac{\tan(\sin x) \tan x \times \sin x}{\sin(\tan x) \tan x \times \sin x}$$
Using $(1)$ and $(2)$, we get
$$=\lim_{x \to 0} \frac{\tan(\sin x) \tan x \times \sin x}{\sin(\tan x) \tan x \times \sin x}$$
$$=\lim_{x \to 0} \frac{\sin x}{\tan x}$$
$$=1$$
Now since we have calculated the limit of the square, we have to take the positive square root (although inconsequential). 
Final answer: $1$.
A: For small $x$, $\sin x\sim\tan x\sim x$, so the numerator and denominator are each asymptotically $$\int_0^x\sqrt{t}dt=\frac23 x^{3/2},$$and the limit is $1$.
