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