Thread: A limit to be solved using L'Hopital's rule

1. A limit to be solved using L'Hopital's rule

Hi,

I would really appreciate any help with finding the following limit:

$\lim _{x \to 0} \frac{\text{arcsin} \sqrt{\text{sin }x}}{\sqrt{2x-x^2}}$

As $x \to 0$, both numerator and denominator $\to 0$, so we can apply the L'Hopital's rule:

$=\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}$

But this is stil an indeterminate form ( $\frac{\infty}{\infty}$), and another application of the L'H rule further complicates the matter...

Thanks for any help!

2. Originally Posted by gusztav
Hi,

I would really appreciate any help with finding the following limit:

$\lim _{x \to 0} \frac{\text{arcsin} \sqrt{\text{sin }x}}{\sqrt{2x-x^2}}$

As $x \to 0$, both numerator and denominator $\to 0$, so we can apply the L'Hopital's rule:

$=\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}$

But this is stil an indeterminate form ( $\frac{\infty}{\infty}$), and another application of the L'H rule further complicates the matter...

Thanks for any help!
Let's look at this a little further

$=\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}$

or

$= \lim _{x \to 0} \frac{1}{\sqrt{1-\sin x}} \frac{\cos x}{\sqrt{\sin x}} \cdot \frac{\sqrt{2x-x^2}}{2-2x}$
$= \lim _{x \to 0} \frac{\cos x}{\sqrt{1-\sin x}(2-2x) } \cdot \frac{\sqrt{2x-x^2}}{\sqrt{\sin x} }$
$= \frac{1}{2} \lim_{x \to 0} \frac{\sqrt{2x-x^2}}{\sqrt{\sin x} }$
$= \frac{1}{2} \sqrt{ \lim_{x \to 0} \frac{2x-x^2}{\sin x } }$
$= \frac{\sqrt{2}}{2}$