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Solve for the limit: $$\lim\limits_{x\to0^+}(\arcsin x)^{\tan(x)}$$

It's $0^0$ expression. I can rewrite this as: $$\lim_{x\to0^+}e^ {\log ({(\arcsin x)^{\tan(x)}})} = \lim_{x\to0^+}e^ {\tan(x) \log ({\arcsin x})}$$ Now it's $e^{- \infty \cdot 0}$. Not much better. Do you have other better ideas how to find this limit?

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    $\begingroup$ $\arcsin x \sim x$ as $x\to 0$, hence $\log (\arcsin x) = \log x + O(1)$, and $\tan x\log (\arcsin x) \to 0$. $\endgroup$ – Daniel Fischer Mar 26 '14 at 10:47
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Every $\infty \cdot 0$ limit can be converted into $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by rearranging into nested fractions. Then you can use l'Hospital. Usually, differentiating the logarithm is a good idea, so I suggest

$$\frac{\log\arcsin x}{1/\tan x}$$

Also, having a tangent to the negative one doesn't make it any worse, because it just becomes its complementary function $\cot x$.

Now you can proceed as usual.

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  • $\begingroup$ To compute these derivatives is waaaay too complicated. Why not rely on simple equivalents? $\endgroup$ – Did Mar 26 '14 at 11:34

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