finding limit w/ fundemental thm: lim[x->0^+](int,0toSin(x))sqrt[tan(t)]dt)/(int,0...

[Blastoise]

finding limit w/ fundemental thm: lim[x->0^+](int,0toSin(x))sqrt[tan(t)]dt)/(int,0...

I was given a few assignments. and I didn't get how to solve one of them

. . . . .\(\displaystyle \large{ \displaystyle \lim_{x \rightarrow 0^+}\, \dfrac{\int_0^{\sin(x)}\, \sqrt{\strut \tan(t)\,}\, dt}{\int_0^{\tan(x)}\, \sqrt{\strut \sin(t)}\, dt} }\)

Im not sure how to approach this. I think it solved with L'Hopital ("0/0") but I'm not sure how I suppose to derivative each of those functions
if there is any trick or way to solve such problems, I would like to know?

thanks

 

[HallsofIvy]

There are two parts of the "Fundamental Theorem of Calculus" but the part you need says that \(\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)\). Here, the upper bound of the two integrals are not "x" but functions of x.

So we need to use the "chain rule". Writing u= g(x), the fundamental theorem becomes \(\displaystyle \frac{d}{du}\int_a^u f(t)dt= f(u)\). Then \(\displaystyle \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}= f(u)\frac{du}{dx}\).

 

[Blastoise]

I think I solve it, if I got it rigt

after two L'Hopital steps I've done, the result of the limit is 1 .

anyone can cofirm?