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
$$\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$$
On
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 $0$.
On
we will use the fact that $\ln(x + a) = \ln x + \frac a x + \cdots$ for $x$ large. $$ \text{ let }y = \frac{(\ln(x+1))^x}{(2x)^{x/3}} .$$ then $$\begin{align}\ln y &= x \ln(\ln(x+1))-\frac x3\left(\ln 2 + \ln x\right)\\ &=x\ln\left(\ln x + \frac 1 x + \cdots\right)-\frac x3\left(\ln 2 + \ln x\right)\\ &=x\left(\ln(\ln x)+\frac 1{x\ln x} +\cdots\right) - \frac x3\left(\ln 2 + \ln x\right)\\ &=-\frac x3\ln x+\cdots \to -\infty \end{align} $$ therefore $$ \lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)=0. $$
In my solution I compare two graphs.
At first: $$\lim_{x\to\infty}\left(\dfrac{\ln(x + 1)}{\sqrt[3]{2x}}\right)^x,$$
as you know $$\lim_{x\to\infty}(q)^x = 0$$ if $$ -1 < q < 1$$
and then I checked is it true that: $$\ln(x + 1)<\sqrt[3]{2x}.$$
I draw 2 graphs for $x>0$
As you can see for $x>0$ inequality is always real so limit is $0$.