<s>[INST] How to solve $\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution with asymptotic approximations, or with variable change(so without Hopital)? Thanks
$$\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$$
 [/INST] Using Taylor expansions:
$$
\ln(x+1) = \ln x + \ln\left(1+\frac{1}{x}\right) = \ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)
$$
and therefore
$$
\ln(x+1)^x = e^{x\ln\left(\ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)\right) }
= e^{x\ln\ln x + x\ln\left(1 + \frac{1}{x\ln x} + o\left(\frac{1}{x\ln x}\right)\right) }
= e^{x\ln\ln x + \frac{1}{\ln x} + o\left(\frac{1}{\ln x}\right) } = e^{x\ln\ln x + o(1)}.
$$
The denominator, however, becomes 
$$
(2x)^{x/3} = e^{\frac{x}{3}\ln(2x)} = e^{\frac{1}{3}x\ln x + \frac{\ln 2}{3}x}
$$
so that overall, the quantity considered is, at infinity,
$$
e^{x\ln\ln x - \frac{1}{3}x\ln x - \frac{\ln 2}{3}x + o(1)}.
$$
But the limit of $x\ln\ln x - \frac{1}{3}x\ln x - \frac{\ln 2}{3}x + o(1)$ is $-\infty$ (can you see why? The term that dominates is $\frac{1}{3}x\ln x$), so by continuity of $\exp$ the limit is $0$.
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