How do I show that $\displaystyle \lim_{x\to\infty}\frac{\arcsin \left(\frac{1}{n^2}\right)}{\arctan \left(\frac{1}{n^2}\right)}=1$?
Thanks!
I've made a correction (in red).
There's no emergency - no need for the hospital:
Let 1/x = t:
$\displaystyle \lim_{t \to 0}\frac{\arcsin \left(t^2\right)}{\arctan \left(t^2\right)}$
Substitute the appropriate Maclaurin series:
$\displaystyle = \lim_{t \to 0}\frac{t^2 + t^6/6 + ....}{t^2 - t^6/3 + ....}$
Divide the top and the bottom by t^2 and take the limit:
$\displaystyle = \lim_{t \to 0}\frac{1 + t^4/6 + ....}{1 - t^4/3 + ....} = 1$