# limit of inverse trigonometric function

I have the following limit that I am trying to solve but apparently I am stuck in doing l'Hôpital's rule and going nowhere so any help would be appreciated

$$\lim_{x\to 0} \frac{\arcsin {x}^2}{(\arcsin x)^2}$$

Thank you

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For small $x$, $\arcsin x\approx x$ (Taylor). This immediately gives you the answer $1$. – Yves Daoust Sep 2 '14 at 6:36

The numerator and denominator both tend to 0. Using L'Hopital's rule once, we get:

$\lim_{x\to 0} \frac{\arcsin {x}^2}{(\arcsin x)^2} = \lim_{x\to 0} \frac{x \sqrt{1-x^2}}{\sqrt{1-x^4} \text{ArcSin}[x]}$

Both the numerator and denominator tend to 0 as x approaches 0, so we use L'Hopital's rule again:

$\lim_{x\to 0} \frac{x \sqrt{1-x^2}}{\sqrt{1-x^4} \text{ArcSin}[x]} = \lim_{x \to 0} \frac{\frac{4 x^4}{\left(1-x^4\right)^{3/2}}+\frac{2}{\sqrt{1-x^4}}}{\frac{2}{1-x^2}+\frac{2 x \text{ArcSin}[x]}{\left(1-x^2\right)^{3/2}}} = 1$

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So using the L'Hospital's Rule two consecutive times will end up solving it right?? I may have been doing some derivative mistake and keep getting 0/0. Thanks a lot! – Moe Oct 13 '12 at 5:00
The derivatives are a bit messy; there is a good chance of making a mistake. You could try testing your answers against what Wolfram Alpha gets, but you should make sure you can do it by hand. – Sean O'Brien Oct 13 '12 at 5:03
what is the derivative of x*sqrt(1-x^2)? is it equal to sqrt(1-x^2)+ [x.(1/2).(1-x^2)^(-0.5).(-2x)] ? – Moe Oct 13 '12 at 5:09
I don't think so. Applying the chain rule you should get $\frac{1}{\sqrt{1-x^2}} + -2 x \frac{-1}{2} \frac{x}{(1 - x^2)^{\frac{3}{2}}} = \frac{1}{\sqrt{1-x^2}}+\frac{x^2}{\left(1-x^2\right)^{3/2}}$ Also, please accept my answer if you are happy with it... :) – Sean O'Brien Oct 13 '12 at 5:14

$$\lim_{x\to 0} \frac{\arcsin {x}^2}{(\arcsin x)^2}$$

$$=\lim_{x\to 0} \frac{\frac{\arcsin {x}^2}{x^2}}{(\frac{\arcsin x}{x})^2}$$

$$=\lim_{x\to 0} \frac{\arcsin {x}^2}{x^2}\cdot \left(\frac{x}{\arcsin x}\right)^2$$

If $\arcsin {x}^2=y,x^2=\sin y$ and $y\to 0$ as $x\to 0$

If $\arcsin {x}=z, x=\sin z$ and $z\to 0$ as $x\to 0$

$$\lim_{y\to 0}\frac{y}{\sin y}\cdot \left(\lim_{z\to 0}\frac{\sin z}{z}\right)^2=1\cdot 1$$

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is this the only way to solve it ? – Moe Oct 13 '12 at 4:49
@Moe, no, you can use L'Hospital's Rule,also. – lab bhattacharjee Oct 13 '12 at 4:51
L'Hospital's Rule keeps giving me 0/0, should it be a derivative mistake I am making? – Moe Oct 13 '12 at 4:54
@Moe, I think you have already found in Sean O'Brien's anser. – lab bhattacharjee Oct 13 '12 at 4:57
yes I did. Thanks – Moe Oct 13 '12 at 4:59