How to find a limit of this? I'm trying to solve this problem,
$\lim_{x\to 0+}(\sin^2(4x))^{\sin^{-1}(2x)}$.
It's indeterminate form so I used L'Hospital's rule but I'm stuck at here;
$\lim_{x\to 0+}\ln(\sin^2(4x))^{\sin^{-1}(2x)}=\lim_{x\to 0+} \frac{2\ln(\sin(4x))}{1/\sin^{-1}(2x)}=\lim_{x\to 0+}\frac{4(\sin^{-1}(2x))^2\sqrt{1-4x^2}}{\tan x} $
Could you help me? Thank you in advance.
 A: Just use that $\lim_{t\to 0^+}t\ln t = 0$ as follows:
You may write
$$\ln\left((\sin^2(4x))^{\sin^{-1}(2x)}\right) =\underbrace{\frac{\sin^{-1} (2x)}{2x}}_{\stackrel{x\to 0^+}{\rightarrow}1}\cdot \underbrace{\frac{4x}{\sin 4x}}_{\stackrel{x\to 0^+}{\rightarrow}1}\underbrace{\sin (4x)\ln (\sin 4x)}_{\stackrel{x\to 0^+}{\rightarrow}0}$$
It follows
$$\lim_{x\to 0+}(\sin^2(4x))^{\sin^{-1}(2x)}= e^0 = 1$$
A: We have that


*

*$\sin^2(4x)=16x^2+o(x^2)$

*$\sin^{-1}(2x)=2x+o(x^2)$
therefore 
$${\sin^2(4x)}^{\sin^{-1}(2x)}=e^{(2x+o(x^2))\log(16x^2+o(x^2))}\to e^0=1$$
A: L'Hospital's rule is not the alpha and omega of limits computations!
It is very simple using equivalents:
We'll first determine the limit of the log:
$$\ln \Bigl(\bigl(\sin^2 4x\bigr)^{\arcsin 2x}\Bigr)=2\arcsin 2x\ln(\sin 4x).$$
Now, we have:


*

*$\arcsin 2x\sim_0 2x$;

*$\sin 4x\sim_0 4x$, so $\ln(\sin 4x)\sim_{0^+} \ln(4x)$.


Therefore
$\;2\arcsin 2x\ln(\sin 4x)\sim_{0^+} 4x\ln(4x)$, which tends to $0$ by a standard high-school limit. By continuity, the limit of the given expression is equal to $1$.
