Here is my attempt to find $\lim_{x\to 0} \arctan (x)^{\frac{1}{x^2}}$ (using Taylor expansions):
$\arctan (x)^{\frac{1}{x^2}} = e^{\ln (\arctan(x)) \frac{1}{x^2}}$
Then, using the Taylor expansion:
$(\arctan{x}) -1 =-1 + x + o(x)$
And composition with $\ln$ (recall $\ln (1 +x) = x + o(x)$):
$\ln(\arctan (x)) = -1 + x + o(x)$
Then, $\ln(\arctan (x)) \frac{1}{x^2} = -\frac{1}{x^2} + \frac{1}{x} + o(\frac{1}{x})$
Then, using the Taylor expansion of $\exp$ (recall $e^x = 1 +x + o(x)$):
$\arctan (x)^{\frac{1}{x^2}} = 1+o(1)$
I am not completely sure about the last step. I feel like it is wrong. Is there another (if it is wrong) way to find this limit ?
Thank you in advance.
Your first step is fine
$$\arctan (x)^{\frac{1}{x^2}} = e^{\ln (\arctan(x)) \frac{1}{x^2}}$$
then recall that $\arctan x \to 0$ and therefore, we are implicitely assuming that $x\to 0^+$, $\ln(\arctan x) \to -\infty$, thus $\ln (\arctan(x)) \frac{1}{x^2} \to -\infty$ and $e^{\ln (\arctan(x)) \frac{1}{x^2}} \to 0$.
We really don't need Taylor expansion in this case.
Moreover this step is wrong
$$\ln(\arctan (x)) = -1 + x + o(x)$$
indeed $\arctan (x)=x + o(x)$ and then $\ln(\arctan (x)) = \ln(x + o(x))$.