Find the following limit,
$\lim_{x\to 0} \frac {\sqrt [5] {x-\tan^{-1} x}}{x^{3/5}}$
Observation: Both top and bottom are 0 for $\lim_{x\to 0}\sqrt [5] {x-\tan^{-1} x}$ and $\lim_{x\to 0}{x^{3/5}}$. Therefore, this seems to hint that I should use the L'Hopital Rule...
So I try it. $\lim_{x\to 0} \frac {\sqrt [5] {x-\tan^{-1} x}}{x^{3/5}}=\lim_{x\to 0}\frac {(\frac{1}{5})({x-\tan^{-1} x})^{-4/5}(1-\frac{1}{x^2+1})}{(\frac{3}{5})x^{-2/5}}=\lim_{x\to 0}\frac {(x)^{2/5}(1-\frac{1}{x^2+1})}{3{({x-\tan^{-1} x})^{4/5}}}$
I notice if I apply L Hopital rule again, I will get back to square one.
After that I tried several ways that included rationalizing the function, etc.
Any hints? Thanks in advance! List them as solutions. I am looking for hints only.