Question

What is the limit as x approaches infinity of `(ln (x))^(1/x)`?

Answer 1

It is quite simple. You must use the fact that
`ln(x) = e^(ln(ln(x)))`
Then, you know that
`ln(x)^(1/x) = e^(ln(ln(x))/x)`
And then, the interesting part happens which could be solved in two ways - using intuition and using maths.

Let us start with intuition part.
`lim_(n->infty) e^(ln(ln(x))/x = lim_(n->infty)e^(("something smaller than x")/x) = e^0 = 1`

Let us think why is that so?
Thanks to continuity of `e^x` function we can move limit:
`lim_(n->infty) e^(ln(ln(x))/x = e^(lim_(n->infty)(ln(ln(x))/x))`

To evaluate this limit `lim_(n->infty)(ln(ln(x))/x)`, we may use de l'Hospital rule which states:
`lim_(n->infty) (f(x)/g(x)) = lim_(n->infty) ((f'(x))/(g'(x)))`

Therefore, when we would count derivatives, we get:

`lim_(n->infty)(ln(ln(x))/x) = lim_(n->infty)(1/(xln(x)))`

As derivatives are `1/(xln(x))` for nominator and `1` for denominator.

That limit is easy to calculate as it is `1/infty` kind of limit which is zero.

Therefore, you see that
`lim_(n->infty) e^(ln(ln(x))/x = e^(lim_(n->infty)(ln(ln(x))/x)) = e^0 = 1`

And it means that `lim_(n->infty) ln(x)^1/x = 1` as well.