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

1 Answer
Dec 23, 2014

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.