Limit of $(\arcsin x)^{\tan x}$ as $x$ tends to zero from the right I came across an interesting limit I could not solve:
$$
\lim_{x \to 0^{+}}\left[\arcsin\left(x\right)\right]^{\tan\left(x\right)}
$$
Given we have not proven l'Hôpital's rule yet, I have to solve it without it. Also, I would rather not use advanced methods such as the taylor series (which yield $x^x$ here).
Squeeze theorem does not (easily?) really help here, nor does the exponent function as far as I see it:
$$
\lim_{x\rightarrow 0+}(\arcsin x)^{\tan\,x} = 
\lim_{x\rightarrow 0+} e^{{\tan(x)}\ln(\arcsin x)}
$$
Here again the exponent is an undefined term $(0 \cdot +\infty)$. Unlike all limits I practiced on however, this logarithm does not tend to $1$, so I don't really see how it cancels out.
Is there an easy solution I am missing?
 A: $$\begin{align} \lim \limits_{x\to0^{+}}(\arcsin(x))^{\tan(x)}&=\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}\\
=e^{\lim_{x\to0^{+}}\frac{\tan(x)}{x}(\arcsin(x)\ln(\arcsin(x)))\frac{x}{\arcsin(x)}
}=1\end{align}$$
Notice that $\lim_{x\to0^{+}}\arcsin(x)\ln(\arcsin(x))=-\lim_{x\to0^{+}}\frac{\ln(\frac{1}{\arcsin(x)})}{{\frac{1}{\arcsin(x)}}}=-\lim_{u\to\infty}\frac{\ln(u)}{u}$
We demonstrate this limit by noticing: $0\le\frac{\ln(u)}{u}=\frac{1}{u}\int_{1}^{u}\frac{1}{t}dt\le\frac{1}{u}\int_{1}^{u}\frac{1}{\sqrt{t}}dt=\frac{2(\sqrt{u}-1)}{u}=2(\frac{1}{\sqrt{u}}-\frac{1}{u})$
for $u$ sufficiently large. So the limit is $0$.
$\lim_{x\to0^{+}}\frac{x}{\arcsin(x)}=\frac{1}{\frac{d(\arcsin(x))}{dx}}|_{x=0^{+}}=\sqrt{1-x^{2}}|_{x=0^{+}}=1$.
Also $\lim_{x\to0^{+}}\frac{\tan(x)}{x}=\lim_{x\to0^{+}}\frac{\sin(x)}{x}\frac{1}{\cos(x)}=1$.
A: $$\lim \limits_{x\to0^{+}}(\arcsin(x))^{\tan(x)}=\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}$$
therefore,
$\tan(x)\ln(\arcsin(x))=\tan(x)\ln\left[x(1+\frac{\arcsin(x)}{x}-1)\right]=$
$$\tan(x)\ln(x)+\tan(x)\frac{\ln(1+\frac{\arcsin(x)}{x}-1)}{\frac{\arcsin(x)}{x}-1} \left (\frac{\arcsin(x)}{x} -1\right)$$ 
the second addend goes to zero as $x \rightarrow 0^{+}$, because 
$(\frac{\arcsin(x)}{x}-1)$ and $\tan(x)$ $\rightarrow 0^{+}$ as $x \rightarrow 0^{+}$ while the central factor tends to 1.
The first addend is equal to
$x\ln(x)\frac{\tan(x)}{x}$ which tends to zero. 
The whole exponent tends to zero, therefore 
$$\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}=e^0=1$$
