<s>[INST] Finding $ \lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}} $ without power series. I have to find: $\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}$
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2}
$
L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?
Edit: I cannot use power series expansion. that is provided on the next semester.
 [/INST] 
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac
{\ln(\frac{\arctan x}{x})} {x^2}$
L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?

By applying L'Hôpital's rule trice and simplifying in each step we obtain:
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{\ln \left( \frac{\arctan x}{x}\right) }{x^{2}}
&=&\lim_{x\rightarrow 0^{+}}\frac{\ln \left( \arctan x\right) -\ln \left(
x\right) }{x^{2}}\tag{1} \\
&=&\frac{1}{2}\lim_{x\rightarrow 0^{+}}\frac{x-\arctan x-x^{2}\arctan x}{
x^{2}\arctan x+x^{4}\arctan x}\tag{2} \\
&=&\lim_{x\rightarrow 0^{+}}\frac{-\arctan x}{4x^{2}\arctan x+x+2\arctan x}
\tag{3}\\
&=&\lim_{x\rightarrow 0^{+}}\frac{-1}{5x^{2}+8x\arctan x+8x^{3}\arctan x+3}
\tag{4}\\
&=&-\frac{1}{3}.\tag{5}
\end{eqnarray*}
ADDED. Explanation:
$\ \:(1)$: $\frac{d}{dx}\left( \ln \left( \arctan x\right) -\ln \left( x\right)
\right) =\dfrac{x-\arctan x-x^{2}\arctan x}{x\left( 1+x^{2}\right) \arctan x}$;
$\ \:(2)$: $\frac{d}{dx}\left( x-\arctan x-x^{2}\arctan x\right) =-2x\arctan x$;
$\ \:(2)$: $\frac{d}{dx}\left( x^{2}\arctan x+x^{4}\arctan x\right) =4x^{3}\arctan
x+x^{2}+2x\arctan x$;
$\ \:(3)$: $\frac{d}{dx}\left( -\arctan x\right) =-\dfrac{1}{1+x^{2}}$;
$\ \:(3)$: $\frac{d}{dx}\left( 4x^{2}\arctan x+x+2\arctan x\right) =\dfrac{5x^{2}+8x\arctan x+8x^{3}\arctan x+3}{1+x^{2}}$. 
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