Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$

Question

I have this simple limit

$$\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}$$

I have solved this with these steps:

$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}=\frac{\ln(1+x^3\cos 3x)(1+\sqrt{1+\cos 3x})}{2x^3(1-\sqrt{1+\cos 3x})\cdot(1-\sqrt{1+\cos 3x})}$$

$$=-\frac 12 \frac{\ln(1+x^3\cos 3x)}{x^3\cos 3x}\cdot (1+\sqrt{1+\cos 3x})\stackrel{x\to 0}{=}-\frac{1+\sqrt 2}{2}$$

My actual problem is to use the asymptotic evaluations: if I use $$\ln(1+x^3\cos 3x)\sim (1+x^3\cos 3x)$$ and $$\sqrt[n]{1+\psi(x)}-1\sim \frac{\psi(x)}{n}$$ where $\psi(x)=\cos 3x$, I obtain

$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}\asymp\frac{(1+x^3\cos 3x)}{2x^3\left(-\frac 12\cos 3x\right)}$$

but is $x\to 0$ this limit diverges. I do not often use asymptotic evaluations but, it may be my fatigue, but I do not find the error at the moment. Thank you for your help.

Answer 1

We have

• $\ln(1+x^3\cos 3x) \sim x^3\cos 3x$

moreover we have $\psi(x) \to 1$ and binomial expansion doesn't apply.

Therefore we obtain

$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}\sim \frac{\cos 3x}{2(1-\sqrt{1+\cos 3x)}} \to\frac1{2(1-\sqrt 2)}$$

Anyway I suggest to proceed with caution for the asymptotic evaluation of limits since it can easily leads to some mistake.

In general it is better use little-o or big-O notation to avoid mistakes.

Answer 2

The fact is that, $\ln(1+x^{3}\cos 3x)\approx x^{3}\cos 3x$, without the additional $1$.