# Thread: Infinite Limit involving Natural Log w/ Exponent

1. ## Infinite Limit involving Natural Log w/ Exponent

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?

2. Originally Posted by gnelly08
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?
It's 1, consider the limit of ln(ln(x)^(1/x))=(1/x)ln(ln(x)).

CB

3. 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 $\displaystyle \color{red}\boxed{\lim_{x\to +\infty}\ln(x)^{\frac{1}{x}}=1}$

4. Alright, I follow you both. =) Thanks for the help.