How to find a limit of this?

Question

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.

Answer 1

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

Answer 2

We have that

• $\sin^2(4x)=16x^2+o(x^2)$

• $\sin^{-1}(2x)=2x+o(x^2)$

therefore

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

Answer 3

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 $1$.