Limit of $\lim_{x \to 0} \arctan(x)^x$ How to calculate $\lim_{x \to 0} \arctan(x)^x$? I was thinking at L'Hospital, but it's no working because I do not have an indeterminate form. I know that arctan(0) is 0, but 0^0 it's a  indeterminate form and I cant figure it out. 
 A: With the properties of $\exp$ and $\ln$ you can write:
$$\arctan(x)^x=\exp\left(\ln\left(\arctan(x)^x\right)\right)=\exp\left(x\cdot\ln\left(\arctan(x)\right)\right)$$
Notice that you need $x>0$ for this, which will not be a problem as you are interested in the limit for $x\to\infty$, so we can assume that $x>0$. Now you can use that $\exp$ is a continuous function and take the limit inside the argument. Can you continue from here?
If instead you are interested in the limit $\lim\limits_{x\downarrow 0}\arctan(x)^x$ (only the one-sided limit is well-defined), we can again assume $x>0$ and arrive at $\lim\limits_{x\downarrow 0} \exp(x\cdot\ln(\arctan(x)))$. With $\exp$ still being continuous you can again take the limit inside. For $\lim\limits_{x\downarrow 0}x\cdot\ln(\arctan(x)))$ you can then use L'Hospital (or use properties of polynomials vs. logarithms).
A: If $x>0$ is near $0$ then $\dfrac x 2 < \arctan x < x,$ so $$ \lim_{x\to0} \left( \frac x 2 \right)^x \le \lim_{x\to0} (\arctan x)^x \le \lim_{x\to0} x^x. $$
A: Hint: For these problems, it is often useful to use that 
$$
\lim_{x\to 0}f(x)=\lim_{x\to 0}\exp (\log f(x)))
$$
A: Observe that for $x\to 0^+$, $\arctan x\sim x +o(x)$
$$
(\arctan x)^x\sim x^x\to1
$$
