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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.

arcilli
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4 Answers4

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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).

Hirshy
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  • The body of the question (as opposed to the title) indicates that the asker is interested in $x \to 0$, and presumably $x \to 0^+$. – Ben Grossmann Sep 03 '17 at 19:16
  • @Omnomnomnom you are absolutely right sir, I will edit my answer accordingly. – Hirshy Sep 03 '17 at 19:17
  • @Hirshy, in what mode to use L'Hospital? It's ok if I try ln(arctan(x)/(1/x) and apply L'Hospital rule with derivative? – arcilli Sep 03 '17 at 19:24
  • @arcilli yes, that is the usual way to apply L'Hospital for problems of this type. – Hirshy Sep 03 '17 at 19:26
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If $x>0$ is near $0$ then $\dfrac x 2 < \arctan x < x,$ so $$ \lim_{x\to0} \left( \frac x 2 \right)^x \le \lim_{x\to0} (\arctan x)^x \le \lim_{x\to0} x^x. $$

Michael Hardy
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Hint: For these problems, it is often useful to use that $$ \lim_{x\to 0}f(x)=\lim_{x\to 0}\exp (\log f(x))) $$

operatorerror
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Observe that for $x\to 0^+$, $\arctan x\sim x +o(x)$ $$ (\arctan x)^x\sim x^x\to1 $$

alexjo
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    Beware! Things are not so simple: $f(x)\sim g(x)$ doesn't necessarily imply $\mathrm e^{f(x)}\sim \mathrm e^{g(x)}$. – Bernard Sep 03 '17 at 19:59