# Simple limit

1. Mar 9, 2009

### Daveyboy

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

lim x->0 (n/log(n))[n1/n-1]

3. The attempt at a solution

I've just been trying to move things around, and use L'Hospitals when appropriate. I haven't been able to see the trick.

Got any hints?

2. Mar 9, 2009

### Focus

Try the algebra of limits

3. Mar 9, 2009

### yyat

Do you mean lim n->infinity?

4. Mar 10, 2009

### Daveyboy

woops, ya as n goes to infinity.

Ive been using L'hospitals on the numerator after multiplying the n through and rewriting in exponential form.

d/dx:[x1/x+1-x] = d/dx:[e(1/x+1)ln(x)-x]= e(1/x+1)ln(x)( -ln(x)/x2+(x+1)/x2) -1
and the bottom goes to 1/x

so the new limit after applying L'Hospital is
e(1/x+1)ln(x)((x+1)-ln(x)/x) -x

which leaves me with a big question mark.

5. Mar 10, 2009

### Daveyboy

also tried looking at it like this.

(n1/n-1)/(log(x)/x)

using L'H on the top and bottom give:

x1/x((1-ln(x)/x2)/(1-log(x))... I want to do something with that log