# Limits help

1. Nov 14, 2007

### Mattofix

1. The problem statement, all variables and given/known data

Find the limit as n tends to infinity of xn = (n^2 + exp(n))^(1/n)

2. Relevant equations

maybe use ( 1 + c/n )^n tends to exp(c)

3. The attempt at a solution

I know that inside the barckets are both inceasing and the 1/n makes it decrease but how do i find out which is stronger and what the limit is?

2. Nov 14, 2007

### JasonRox

Try finding the limit of the ln of the function.

Tip: ln(a^b) = b ln(a)

3. Nov 14, 2007

### Mattofix

i already tried that and i cant see how it helps, (1/n)ln(n^2 + exp(n)) has the same problem...

4. Nov 14, 2007

### Dick

Try l'Hopital's rule, if you know that.

5. Nov 14, 2007

### Mattofix

i have a feeling im not allowed to use it, is there another way?

6. Nov 14, 2007

### Dick

What you really need to know is that n^2/exp(n)->0 as n->infinity. There are a variety of ways to show that - try to think of one. Once you done that then ln(exp(n)+n^2)=ln(exp(n)*(1+n^2/exp(n))=ln(exp(n))+ln(1+n^2/exp(n)) etc.

7. Nov 14, 2007

### Kummer

$$(n^2+e^n)^{1/n} = e(1+n^2/e^n)^{1/n}$$ now it is trivial but the fact that $$(1+x_n/n)^{1/n} \to e^x$$ if $$x_n\to x$$.