Help with the limit

[koutos]

 

[daon]

Hint: ln(n) <= n and for n>2, 1 < ln(n)

 

[koutos]

Thanks for the hint, it was very useful.

I solved it, i think.

Here is my solution:

Am i right?

 

[daon]

you got it! I would make mention of the squeeze theorem. Words in proofs aren't bad things

 

[lookagain]

Am i right?

I see the following to finish part of the last two lines:


\(\displaystyle \sqrt[n]{1} \le \sqrt[n]{ln(n)} \le \sqrt[n]{n}\)


\(\displaystyle \lim_{n \to \infty}\sqrt[n]{1} = \lim_{n \to \infty}\sqrt[n]{(n)} \rightarrow \lim_{n \to \infty}\sqrt[n]{ln(n)} = 1\)


The middle expression in your bottom line of this quote box looks wrong to me.