# Limit with integral

1. Mar 29, 2009

### rey242

Limit with integrel

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

2. Relevant equations

3. The attempt at a solution
I tried to take
x=ln(u)
dx=du/u
and solve the integral but I keep getting stuck at the int(cos(u)/u.)
Can anyone help me out here?

2. Mar 29, 2009

### Dick

Re: Limit with integrel

You can't take the limit x->infinity. x is a dummy integration variable. You must mean n->infinity. Right?

3. Mar 29, 2009

### rey242

Re: Limit with integrel

I meant N.
Sorry, its a force of habit.

4. Mar 29, 2009

### Dick

Re: Limit with integrel

Ok, then as n->infinity then 1/n -> 0. You don't actually have to do the integral to know what the limit is, because cos(e^x) is bounded. What's the limit of the integral of a bounded function between 0 and 1/n as n->infinity?

5. Mar 29, 2009

### rey242

Re: Limit with integrel

so the integral would be zero as n approaches inf. since the limits of integration go from zero to zero, right?

6. Mar 29, 2009

### Dick

Re: Limit with integrel

No, it's not just the limits of integration, you have to think about how the function you are integrating behaves. -1<=cos(e^x)<=1. Agree? So what limits can you make for the integral from 0 to 1/n of cos(e^x)? What happens as n->infinity?

7. Mar 29, 2009

### JG89

Re: Limit with integrel

I'd try the MVT.