1. ## Limits

Can someone help me with this problem:

lim (n^2 + n)^(1/n) = 1;

In (b) you may assume that n^1/n = 1.

I know how to solve these when it's a number to the power of n but not when it's n to a power.

2. $\displaystyle (n^2 + n)^{1/n} = [n(n+1)]^{1/n} = n^{1/n} (n+1)^{1/n}$.

Can you show that $\displaystyle (n+1)^{1/n}$ goes to 1 as n goes to infinity?

3. Originally Posted by Mathsnewbie
Can someone help me with this problem:

lim (n^2 + n)^(1/n) = 1;

In (b) you may assume that n^1/n = 1.

What does this mean??

I know how to solve these when it's a number to the power of n but not when it's n to a power.

If you already know $\displaystyle \sqrt[n]{n}\xrightarrow [n\to\infty]{}1$ , then it's easy: $\displaystyle \sqrt[n]{n^2+n}=\sqrt[n]{n}\sqrt[n]{n+1}\xrightarrow [n\to\infty]{}1\cdot 1=1$ .

Tonio

4. Originally Posted by JG89
$\displaystyle (n^2 + n)^{1/n} = [n(n+1)]^{1/n} = n^{1/n} (n+1)^{1/n}$.

Can you show that $\displaystyle (n+1)^{1/n}$ goes to 1 as n goes to infinity?
I can't, I know how to do some by sandwich principle but I couldn't get either of these to work.

5. Originally Posted by Mathsnewbie
I can't, I know how to do some by sandwich principle but I couldn't get either of these to work.

$\displaystyle \sqrt[n]{a}\xrightarrow [n\to\infty]{}1\,,\,\,\forall a>0$ , so just check that $\displaystyle 1\leq\sqrt[n]{n+1}\leq \sqrt[n]{2n}=\sqrt[n]{2}\,\sqrt[n]{n}$ ...

Tonio

6. Thank you, I managed to complete