# Thread: [SOLVED] Limit we've seen a number of times

1. ## [SOLVED] Limit we've seen a number of times

I should just look this up, but I can't figure out how to do a search for it. It's cropped up a couple of times, but I don't recall the answer, much less how to find it.

Anyway, here it is
$\lim_{n \to \infty} \sqrt[n]{n}$
(I'm remembering something about taking the log of the limit perhaps?)

-Dan

1

I will try to remember how to get this!

A little later...

Rewrite this as:

e^(ln(n)/n)

Then use L'Hostpital on ln(n)/n.

Lim e^(1/n) = 1
n→∞

3. Originally Posted by spammanon
1

I will try to remember how to get this!

Rewrite this as:

e^(ln(n)/n)

Then use L'Hostpital on ln(n)/n.

Lim e^(1/n)
n→∞
Aaah! I'd forgotten about that little trick. Thank you. (Yay! I remembered the answer correctly!)

-Dan

4. Do it without L'Hopital.

$\ln n = \int_1^n \frac{dx}{x}$.

But, $0\leq \frac{1}{x} \leq \frac{1}{\sqrt{x}}$ for $x\in [1,n]$ for all $n\in \mathbb{Z}^+$.

Thus, $0\leq \int_1^n \frac{dx}{x} \leq \int_1^n \frac{dx}{\sqrt{x}} = 2\sqrt{n} - 2 \leq 2\sqrt{n}$.

Thus, $0 \leq \frac{\ln n}{n} \leq \frac{2\sqrt{n}}{n} = \frac{2}{\sqrt{n}}$.

Now use the Squeeze Theorem.