This is an exam problem from my analysis 1 course, and I can't find a way to solve it.
$$\lim_{x\to 0} \frac {\arcsin(x) \sqrt{\sin(x)}}{\sqrt{2x-x^2}}$$
So far I have tried applying L'Hopital's rule since, as x approaches zero, we get $\frac{0}{0}$ as a result. Eventually I got:
$$ \lim_{x\to 0} \frac{\frac{\sqrt{\sin (x)}}{\sqrt{1-x^2}}+\frac{\arcsin(x)\cos(x)}{2\sqrt{\sin(x)}}}{\frac{1-x}{\sqrt {2x-x^2}}}$$
Fast forward few steps, after trying to get rid of the double fractions, I got
$$ \lim_{x\to 0}\frac{\sqrt{2x-x^2}(2\sin(x)+\sqrt{1-x^2}\arcsin(x)\cos(x))}{2(1-x)(\sqrt{1-x^2})(\sqrt{\sin (x)})} $$
And I get again $\frac{0}{0}$ when I let x approach zero. I honestly doubt I should apply L'Hopital again here, and since I can't see any other way around I am asking for help or a clue how to solve this problem.
Thanks in advance
Best Answer
The trick is to keep multiplying and dividing by powers of $x$ like so
$$\lim_{x\to0^+}\frac{\arcsin x \sqrt{\sin x}}{\sqrt{2x-x^2}} = \lim_{x\to0^+}\frac{\arcsin x}{x}\cdot \frac{x}{\sqrt{2x-x^2}}\cdot \sqrt{\frac{\sin x}{x}}\cdot\sqrt{x}$$
$$\lim_{x\to0^+} \frac{\arcsin x}{\sin(\arcsin x)}\cdot \frac{\sqrt{x}}{\sqrt{2-x}}\cdot\sqrt{\frac{\sin x}{x}}\cdot \sqrt{x} \to 1 \cdot 0 \cdot 1\cdot 0 = 0$$
by using the known limits (manipulating $\frac{\sin x}{x} \to 1$) for each of the others. But the double-sided limit does not exist.