Find the limit of a sequence?

Question

Hi, thanks for help ahead of time!

Show that the lim{ n^(1/[n-1]) } as n->infinity equals 1.

Hint: use that lim{log(n)/(n-1)}=0

I know how to prove the hint (L'Hospitals Rule). I also know that by taking the log of n^(1/[n-1]) I get the desired form log(n)/(n-1). I am not sure what step allows me to take the log of the original sequence, though. Any help would be much appreciated.

Answer 1

if you want, you may set y = n^(1/ (n-1) ).  Then take the logarithm of both sides.

However, I personally prefer writing it like this:

n^(1/ (n-1) ) = (e^log(n))^(1/ (n-1) )

= e^(log(n) / (n-1) )

Then you can see that we're actually taking the limit of the exponent.

Answer 2

lim { n^(1/[n-1])} = e ^lim ln {n^(1/[n-1]) = e^0 = 1....if g(x) is continuous  at  x = a and if lim f(x) = a then lim g(f(x)) = g ( lim f(x) ) = g(a)