How to find a limit of this?

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 $$\lim_{x\to 0+}(\sin^2(4x))^{\sin^{-1}(2x)}= e^0 = 1$$

We have that

• $\sin^2(4x)=16x^2+o(x^2)$

• $\sin^{-1}(2x)=2x+o(x^2)$

therefore

$${\sin^2(4x)}^{\sin^{-1}(2x)}=e^{(2x+o(x^2))\log(16x^2+o(x^2))}\to e^0=1$$