How do I find such a limit: $\lim_{x \to 0}\frac{\int_0^{\sin x} \sqrt{\tan t} \ dt}{\int_0^{\tan x} \sqrt{\sin t} \ dt}$? How do I find the limit
$$
\lim_{x \to 0}\frac{\int_0^{\sin x} \sqrt{\tan t} \,\mathrm dt}{\int_0^{\tan x} \sqrt{\sin t} \,\mathrm dt}?
$$
The functions in numerator and denominator are a mirror image to each other. That's why I believe it is possible to merge them together under the limit. However, I know that I can go with the limit "under the integral" and I'm not sure how to merge those 2 integrals.
Any help would be much appreciated.
 A: I understand the question as follows:
Determine the limit
$$\lim_{x \to 0}\frac{\int_0^{\sin x} \sqrt{|\tan \xi|}d\xi}{\int_0^{\tan x} \sqrt{|\sin \xi|}d\xi}. $$
Let $f(x) = \int_0^{\sin x} \sqrt{|\tan \xi|}d\xi$ and $g(x) = \int_0^{\tan x} \sqrt{|\sin \xi|}d\xi$. Observe that
$$1) \ f'(x)=\cos(x)\sqrt{|\tan (\sin x)|}, \ \mathrm{and} \ g'(x) = \frac{1}{\cos^2x}\sqrt{|\sin(\tan x)|}$$ by the fundamental theorem of calculus. Note that we have the following Taylor expansions around $x=0:$
$$2) \ \tan(\sin x) = x + \mathcal{O}(x^3), \ \ \sin(\tan x) = x + \mathcal{O}(x^3). $$
Using 1) and 2), it follows that we have the limit
$$3) \ \lim_{x \to 0}\frac{f'(x)}{g'(x)} = \lim_{x \to 0}\frac{\cos^3(x)\sqrt{x+\mathcal{O}(x^3)}}{\sqrt{x+\mathcal{O}(x^3)}}= 1.$$
We also note that
$$4) \ \lim_{x \to 0}f(x) = \lim_{x \to 0}g(x)=0.$$
By l'Hôpital's rule, it now follows from 3) and 4) that
$$\lim_{x \to 0}\frac{\int_0^{\sin x} \sqrt{|\tan \xi|}d\xi}{\int_0^{\tan x} \sqrt{|\sin \xi|}d\xi} = \lim_{x \to 0}\frac{f(x)}{g(x)} = \lim_{x \to 0}\frac{f'(x)}{g'(x)}=1. $$
A: Since:
$$\lim_{x\to0}\underbrace{\int\limits_0^{\sin x}\sqrt{\tan t}\,\mathrm dt}_{f(x)}=\lim_{x\to0}\underbrace{\int\limits_0^{\tan x}\sqrt{\sin t}\,\mathrm dt}_{g(x)}=0$$
We can use L'Hopital's rule and say that:
$$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
Now by FTC we get that:
$$f'(x)=\left.\sqrt{\tan t}\right|_{t=0}^{\sin x}=\cos x\sqrt{\tan(\sin x)}$$
similarly:
$$g'(x)=\sec^2 x\sqrt{\sin(\tan x)}$$

If we define:
$$L=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
It follows that:
$$L^2=\lim_{x\to0}\frac{\tan(\sin x)}{\sin(\tan x)}$$

Looking at both functions, you will see that around zero:
$$\tan(\sin x)\sim\sin(\tan x)\sim x$$ and so we are left with:
$$L^2=\lim_{x\to0}\frac{\cos^2x}{\sec^4x}=\lim_{x\to0}\cos^2x=1$$
