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 ) .

  • $\begingroup$ Maybe, instead of solving the integrals you should do some approximation. I would guess the result is $1$ -- for small $x$ the upper bounds are approximately $x$ thus functions are integrated with small $t$'s only and thus both approximately equal to $\sqrt{t}$ $\endgroup$
    – guest
    Mar 13, 2020 at 15:54
  • $\begingroup$ I have the following answers : a)l=-1 b) l=0 c)l=1 d) l=2 e) l= +infinity f) l= - infinity $\endgroup$
    – Jon9
    Mar 13, 2020 at 15:57

2 Answers 2


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 $\sim\int_0^x\sqrt{t}dt=\frac23x^{3/2}$.

  • $\begingroup$ Awesome ! Thank you ! I also have a question . What you did is to differentiate under the integral sign in order to remain only with the functions, right ? $\endgroup$
    – Jon9
    Mar 13, 2020 at 16:19
  • $\begingroup$ @Jon9 Oh, no, nothing like that. I just used the chain rule:$$\frac{d}{dx}\int_a^{f(x)}g(t)dt=f^\prime(x)\frac{d}{df}\int_a^fg(t)dt=f^\prime(x)g(f(x)).$$ $\endgroup$
    – J.G.
    Mar 13, 2020 at 16:41
  • 1
    $\begingroup$ Now i see . It becomes quite simple after you apply the chain rule .( i thought the problem should have trick otherwise it would have been too difficult) . Thank you again for your help ! $\endgroup$
    – Jon9
    Mar 13, 2020 at 16:46

Use Taylor series for the integrands; integrate termwise. Apply the bounds and repeat. You should have $$\text{numerator}=\frac{2 x^{3/2}}{3}-\frac{5 x^{7/2}}{42}+O\left(x^{11/2}\right)$$ $$\text{denominator}=\frac{2 x^{3/2}}{3}+\frac{13 x^{7/2}}{42}+O\left(x^{11/2}\right)$$ Now, long division to get as a result $$1-\frac{9 x^2}{14}+O\left(x^4\right)$$ which shows the limit and how it is approached.

Using numerical integration with $x=\frac \pi{24}$, the result is $0.989012$ while the above gives $0.988985$.


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