One side limit $\,\big(1+\arcsin\left(4x\right)\big)^{\frac{1}{-3x}}$ Can you help me with this limit? 
It is one side limit. Please if it is possible tell me some trick(formula) or something like that or where I can find it. Thanks a lot.
$$ \lim_{x\to0^-}{\big(1+\arcsin\left(4x\right)\big)^{\frac{1}{-3x}}}$$ 
 A: $$ \lim_{x\to0^-}{\big(1+\arcsin\left(4x\right)\big)^{\frac{1}{-3x}}}$$ 
$$ =\lim_{\arcsin\left(4x\right)\to 0}[{\big(1+\arcsin\left(4x\right)\big)^{\frac{1}{\arcsin\left(4x\right)}}]^{\frac{\arcsin\left(4x\right)}{-3x}}} $$
$$ =e^{\lim_{x\to0^-}\frac{\arcsin\left(4x\right)}{-3x}} $$
$$ =e^{-\frac{4}{3}\lim_{x\to0^-}\frac{\arcsin\left(4x\right)}{4x}} $$
$$ =e^{-\frac{4}{3}}$$
A: $$\lim_{x\to 0}\left(1+\sin^{-1}\left(4x\right)\right)^{-\frac{1}{3x}}=$$
$$\lim_{x\to 0}\exp\left(\ln\left(\left(1+\sin^{-1}\left(4x\right)\right)^{-\frac{1}{3x}}\right)\right)=$$
$$\lim_{x\to 0}\exp\left(-\frac{\ln\left(1+\sin^{-1}\left(4x\right)\right)}{3x}\right)=$$
$$\exp\left(\lim_{x\to 0}-\frac{\ln\left(1+\sin^{-1}\left(4x\right)\right)}{3x}\right)=$$
$$\exp\left(-\frac{1}{3}\left(\lim_{x\to 0}\frac{\ln\left(1+\sin^{-1}\left(4x\right)\right)}{x}\right)\right)=$$
$$\exp\left(-\frac{1}{3}\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\ln\left(1+\sin^{-1}\left(4x\right)\right)}{\frac{\text{d}}{\text{d}x}x}\right)\right)=$$
$$\exp\left(-\frac{1}{3}\left(\lim_{x\to 0}\frac{\frac{4}{\sqrt{1-16x^2}(1+\sin^{-1}(4x))}}{1}\right)\right)=$$
$$\exp\left(-\frac{1}{3}\left(\lim_{x\to 0}\frac{4}{\sqrt{1-16x^2}\left(1+\sin^{-1}\left(4x\right)\right)}\right)\right)=$$
$$\exp\left(-\frac{1}{3}\left(\frac{4}{\sqrt{1-16\cdot 0^2}\left(1+\sin^{-1}\left(4\cdot 0\right)\right)}\right)\right)=$$
$$\exp\left(-\frac{1}{3}\left(\frac{4}{1}\right)\right)=$$
$$\exp\left(-\frac{4}{3}\right)=e^{-\frac{4}{3}}=\frac{1}{\sqrt[3]{e^4}}$$
A: We use the usual trick of turning a $1^\infty$ indeterminate form into $\frac{0}{0}$ by using the fact the exponential and natural logarithmic functions are inverses of each other.  (Note that $\exp x = e^x$ -- it can be handy to use when we have an ugly exponent.)
\begin{align*}
\lim_{x\rightarrow 0^{-}}(1+\arcsin(4x))^{\frac{1}{-3x}}&=\exp\left(\lim_{x\rightarrow 0^{-}}\ln(1+\arcsin(4x))^{\frac{1}{-3x}}\right)\\
&=\exp\left(-\lim_{x\rightarrow 0^{-}}\frac{\ln(1+\arcsin(4x))}{3x}\right)\\
\end{align*}
Now we use L'Hopital's rule (lots of chain rule):
\begin{align*}
&=\exp\left(-\lim_{x\rightarrow 0^-}\left(\frac{1}{1+\arcsin(4x)}\cdot\frac{1}{\sqrt{1-(4x)^2}}\cdot 4\cdot\frac{1}{3}\right)\right)
\end{align*}
And now we can substitute $x=0$:
\begin{align*}
&=e^{-\frac{4}{3}}.
\end{align*}
A: $$\lim\limits_{x\to 0^-}\left(1+\arcsin\left(4x\right)\right)^{-\frac{1}{3x}}$$
$$=\lim\limits_{x\to 0^-}\exp\left(\ln\left(1+\arcsin\left(4x\right)\right)^{-\frac{1}{3x}}\right)$$
$$=\lim\limits_{x\to 0^-}\exp\left(-\frac{1}{3x}\ln\left(1+\arcsin\left(4x\right)\right)\right)$$
$$=\exp\left(-\frac13\lim\limits_{x\to 0^-}\frac{\ln\left(1+\arcsin\left(4x\right)\right)}{x}\right)$$
Let $h=\arcsin(4x)$, then
$$\exp\left(-\frac43\lim\limits_{h\to 0}\frac{\ln\left(h+1\right)}{\sin h}\right)$$
$$=\exp\left(-\frac43\lim\limits_{h\to 0}\frac{\frac{d}{dh}\ln\left(h+1\right)}{\frac{d}{dh}\sin h}\right)$$
$$=\exp\left(-\frac43\lim\limits_{h\to 0}\frac{\frac1{h+1}}{\cos h}\right)$$
$$=\exp\left(-\frac43\right)$$
