Question

Why is the limit of n^(1/n) as n approaches infinity 1?

Answer 1

3,10,17,24, ...
what will be the 100th number in the pattern?

Answer 2

3,10,17,24, ...
what will be the 100th number in the pattern?

Answer 3

Well we're approaching this weird indeterminate form \(\large\rm \infty^0\).
It's sort of hard to tell which is happening faster though, right?
Is the exponent approaching 0 fast enough to kill the base?

We can use that old exponent log trick, leading into L'Hopital's Rule to determine what's going on.

Answer 4

Recall this little trick,\[\large\rm e^{\ln\left(\color{orangered}{x}\right)}=\color{orangered}{x}\]Exponential base e and log base e are inverse operations of one another,
so taking their composition gives us the argument back.

We want to apply this trick in reverse,\[\large\rm \color{orangered}{\lim_{n\to\infty}n^{1/n}}=e^{\ln\left(\color{orangered}{\lim_{n\to\infty}n^{1/n}}\right)}\]

Answer 5

Pass the log into the limit using limit laws,\[\Large\rm e^{\lim_{n\to\infty}\ln\left[n^{1/n}\right]}\]And from there, let's ignore the base for now.
Just pay attention to what's going on up in the exponent.\[\large\rm \lim_{n\to\infty}\ln\left[n^{1/n}\right]\]

Answer 6

Log rule allows us to bring the 1/n out in front of the log, ya?
And then some stuff from there.

Answer 7

man this calc II stuff is killer

Answer 8

they werent kidding when they said this would be a tough class. I see what your saying, thanks for explaining it so thoughtfully zep!

Answer 9

Do you understand how to finish it up from there? :O