Evaluating $\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)} $

Question

I used all the trig and log tricks but still can't compute this limit

$$\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}.$$

I tried the following:

\begin{align*}\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)} &= \lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(x)} \\ &= \lim_{x\to0}\frac{\ln\left(1+4\sin^2(x)\cos^2(x)\right)}{\sin^2(x)} .\end{align*}

I tried to substitute $\sin^2(x)$ but still wandering.

Answer 1

Hint:- The Maclaurin Expansion of $\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-....\,,-1<x\leq 1$

This gives $\ln(1+x)=x+O(x^{2})$ . Where $O(.)$ is the big O notation

Can you use this now? . Try replacing $x$ with $4\sin^{2}(x)\cos^{2}(x)$.

Another hint:- It is well known that $$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$$.

Then $$\lim_{x\to 0} 4\cos^{2}(x)\frac{\ln(1+4\sin^{2}(x)\cos^{2}(x))}{4\sin^{2}(x)\cos^{2}(x)}$$. Can you now see the similarity between the two expressions above? . If yes then what can you say by product rule of limits?

Answer 2

We know that (by asymptotic relations): $$\ln(1+f(x)) \sim f(x)\;\;\; x \to \alpha \;\; f(x) \to 0$$ Also: $$\sin(x) \sim t \;\; x \to 0$$ And: $$\cos(x) \sim 1 - \frac{1}{2}x^2 \;\; x \to 0$$

So, your limit is going to be: $$\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}\;\;\sim \;\;\lim_{x\to0}\frac{\sin^2(2x)}{1-\cos^2(x)}\;\;\lim_{x\to0}\frac{4x^2}{1-\cos^2(x)}=\lim_{x\to0}\frac{4x^2}{(1+\cos(x))\cdot(1-\cos(x))}\;\; \lim_{x\to0}\frac{4x^2}{2\cdot\frac{1}{2}x^2}=4$$

Answer 3

Just using the standard limit $\lim_{t\to 0}\frac{\ln (1+t)}{t} = 1$ you get

\begin{eqnarray*}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)} & = & \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)}\cdot \frac{\sin^2(2x)}{\sin^2 x} \\ & = & 4\cos^2 x \cdot \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)} \\ & \stackrel{x\to 0}{\longrightarrow} & 4\cdot 1 \cdot 1 = 4 \end{eqnarray*}