<s>[INST] Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$ 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.
 [/INST] 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.
</s>