Solve : $$l=\lim\limits_{x \to 0,x>0}\frac{\int\limits_0^{\sin x} \sqrt{\tan t} \; dt}{\int\limits_0^{\tan x} \sqrt{\sin t}\; dt}$$
So, the main problem that I have is to evaluate $\int\limits_0^{\tan x} \sqrt{\sin t}\; dt$. I was looking to see exactly how to deal with this integral , but it seems like this integral requires college level skills (which I don't possess).
This problem is from a highschool textbook, so I thought that there is some trick that can be applied to it without having to know stuff from college. The main problem is with this integral because I've managed to solve the first one.
Also, it seems like the limit itself will be a problem once i will deal with the second integral (what I' ve tried to do to the second integral is to substitute $u=\sqrt{\sin t}$ followed by $\sin x=u^2 \implies \cos x\;dx= 2u\; du \implies dx=\frac{ 2u}{\sqrt{1-u^4} }\;du$ but I am still getting stuck ) .
Best Answer
By L'Hôpital's rule the limit is$$\lim_{x\to0^+}\frac{\cos x\sqrt{\tan\sin x}}{\sec^2x\sqrt{\sin\tan x}}=\sqrt{\lim_{x\to0^+}\frac{\tan\sin x}{\sin\tan x}}=1.$$In particular,$$\lim_{x\to0^+}\frac{\tan\sin x}{x}=\lim_{x\to0^+}\frac{\tan\sin x}{\sin x}\lim_{x\to0^+}\frac{\sin x}{x}=\lim_{y\to0^+}\frac{\tan y}{y}\lim_{x\to0^+}\frac{\sin x}{x}=1\times1=1$$etc., so$$\lim_{x\to0^+}\frac{\tan\sin x}{\sin\tan x}=\frac{\lim_{x\to0^+}\frac{\tan\sin x}{x}}{\lim_{x\to0^+}\frac{\sin\tan x}{x}}=\frac11=1.$$If L'Hôpital's rule is unavailable to you, note $\sin x\sim x$ and $\tan x\sim x$, ad so each integral $\sim\int_0^x\sqrt{t}dt=\frac23x^{3/2}$.