Alright, I'm not the greatest mathematician so I thought I would ask for help on this one, it seems a bit more complicated than I'm used to.
as limit x approaches +infinity (ln x)^(1/x)
Any thoughts?
Hello,
There are several methods, I suggest you this :
$\displaystyle \ln(x)^{\frac{1}{x}}=\exp\left(\frac{\ln(\ln(x))}{ x}\right)$
We want to find the limit of $\displaystyle \frac{\ln(\ln(x))}{x}$ when $\displaystyle x\to \infty$.
Take $\displaystyle y=\ln(x)$ then $\displaystyle \frac{\ln(y)}{e^y}$.
And for $\displaystyle y\to \infty$ we have $\displaystyle 0\le \frac{\ln(y)}{e^y}\le \frac{y-1}{e^y}\longrightarrow 0$
So $\displaystyle \lim_{y\to \infty} \frac{\ln(y)}{e^y}=0$ and therefore $\displaystyle \color{red}\boxed{\lim_{x\to +\infty}\ln(x)^{\frac{1}{x}}=1}$