# Limit of e^(i*x) at infinity?

1. Nov 30, 2013

### Typhon4ever

1. The problem statement, all variables and given/known data
delta dirac function(x) * e^(-i*x) at ∞ and -∞

2. Relevant equations
delta dirac(x)*e^(i*x)

3. The attempt at a solution
I'm wondering how e^(i*x) looks like at infinity/-infinity. I know it has some sort of oscillating property i.e e^(pi*i)=e^(3pi*i). The problem is I'm trying to find the delta dirac function * e^(-i*x) at infinity and negative infinity but I don't know how to evaluate the exponential. I know the delta function should be 0 in both cases but I don't know if the exponential will turn out to be indeterminate or something else. I should get 0 from the whole equation.

2. Nov 30, 2013

### vela

Staff Emeritus
3. Nov 30, 2013

### Typhon4ever

Ok. Using Euler's if I'm not mistaken just tells me the limit doesn't exist. What do I do then to evaluate this problem?

4. Nov 30, 2013

### Mandelbroth

How are you defining the Dirac delta? What does that imply about the product?

5. Nov 30, 2013

### Typhon4ever

I'm not sure what you mean by how I'm defining the dirac delta.

6. Nov 30, 2013

### Mandelbroth

Well, what does the Dirac delta evaluate to at any nonzero $x\in\mathbb{R}$?

I'm assuming your "*" means multiplication, correct?

7. Nov 30, 2013

### Typhon4ever

Yes "*" means multiplication. Anywhere except the origin is 0.

8. Nov 30, 2013

### vela

Staff Emeritus
If you're referring to the limit $\displaystyle \lim_{x \to \infty} e^{ix}$, then you're correct. That limit doesn't exist because $e^{ix}$ doesn't converge to a single value. But that's not the limit you were asking about. You should cogitate on the implication of your (correct) answer to Mandelbroth's question.

9. Nov 30, 2013

### Typhon4ever

So I have the delta function at infinity which is 0 * a non converging exponential=...0?

10. Nov 30, 2013

### vela

Staff Emeritus
Yes, that's the right answer, but you're not thinking about this correctly. Since you're interested in the limit as x goes to +∞, you can assume x>0. For any such value of x, $\delta(x)e^{ix} = 0\times e^{ix} = 0$. Now you take the limit of this as x goes to +∞.
$$\lim_{x \to \infty} \delta(x)e^{ix} = \lim_{x \to \infty} 0 = 0.$$ The behavior of $e^{ix}$ doesn't matter.

11. Nov 30, 2013

### Typhon4ever

Oh, ok. Thank you!