Let $\textit{f} :[-1,1] \rightarrow \mathbb{R}$ continuous on $[-1,1]$

I need to prove that $$\lim_{x \rightarrow{1}^{-}} \int_{-x}^{x}\frac { f(t)}{\sqrt { 1-{ t }^{ 2 } }}dt$$


But I have no clue of the value of the limit (I know it has to be finite)

And since there is no information on the sign of $f$, I cannot use any "it is an increasing and bounded function" argument

Thanks for your help


Let $t=\sin a$. The function $f(\sin a)$ is continuous in $[-\arcsin x, \arcsin x]$. The limit exists. You can use Fundamental theorem of calculus. http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus 'first part'


One can guess the limit is $\int_{-1}^1\frac{f(t)}{\sqrt{1-t^2}}dt$. First, this integral makes sense because $f$ is bounded on $[-1,1]$ and $\int_0^1\frac 1{\sqrt s}ds$ is convergent. Then, to show that it's indeed the limit, use continuity of $f$ at $1$ and $-1$.


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