how to find the lim of $(1+\arcsin x) ^{\cot x} $ as $x$ goes to $0$? $$\lim_{x\to 0} \ln (1+\arcsin x) ^{\cot x}=\lim_{x\to 0} \cot x \ln (1+\arcsin x)=\lim 
\frac{(1+\arcsin x)}{\tan x}$$ from l'hospital 
$$ \lim_{x\to 0}\frac{ \frac{1} {\sqrt{1-x^2}}} {\sec^2 x} $$ is it true? 
 A: Hint : You can write the expression as
$$(1+\arcsin(x))^{\frac{1}{\arcsin(x)} \times {\frac{\arcsin(x)}{\tan(x)}}}$$
and $\frac{\arcsin(x)}{\tan(x)}$ tends to $1$, so you finally get $e$ as the limit.
A: Use the good old taylor series:
$$A:= (1+\arcsin x) ^{\cot x} \implies \ln(A) = \ln((1+\arcsin x)) \cot(x) $$
If $x$ is small, which is true in this case because $x \to 0$, I can approximate this as 
$$\ln(A) = \ln(1+x)\cot(x) + O(x^2) = \ln(1+x) \frac{1}{x} + O(x^2) = \frac{x}{x} + O(x^2) =1 + O(x^2)$$
$$\implies \lim_{x \to 0} \, \ln(A) = 1 \implies A= e $$
If you don't like the big $O$ notation above you can change equal signs with approximation signs and live the $O(x^2)$ terms out. Since the approximation gets better and better as $x \to 0$ it won't be a problem to do so.
A: $$\begin{align}\lim_{x\to 0} (1+\arcsin x) ^{\cot x}&=\lim_{x\to 0} e^{{\cot x}\ln(1+\arcsin x) }\\&=\exp\lim_{x\to 0} ({\cot x}\ln(1+\arcsin x) )\\&=\exp\lim_{x\to 0} ({\cot x\arcsin x}(\ln(1+\arcsin x))/(\arcsin x) )\\&=\exp\lim_{x\to0}({\cot x\arcsin x} )\tag{$\because \lim_{x\to0}\frac{\ln(1+x)}x=1$}\\&=\exp\lim_{x\to0}\left(\frac{\arcsin x}{\tan x }\right)\\&=\exp\lim_{x\to0}\left(\frac{1/\sqrt{1-x^2}}{\sec^2x}\right)\tag{L'Hospital}\\&=\large \rm e
\end{align}$$
A: $\lim_{x\to 0} (1+\arcsin x)^{\cot x}=\lim_{x\to 0}  (1+ x + \cdots)^{\dfrac{1}{x} + \cdots}= e$ now take the logarithm we get 
$$\lim_{x\to 0}\ln  (1+\arcsin x)^{\cot x}= 1 .$$
