# Limit of a complex exponential

1. May 26, 2014

### Waxterzz

Hi,

The two terms should vanish at infinity according to the Quantum textbook of Griffiths, but I don't see how?

I mean a complex exponential is a periodic function so how can it vanish at infinity?

If you split up the first term

exp(ikx) * exp(-ax)

Take the limit of infinity of this expression

lim exp(ikx) is not defined
limit exp(-ax) turns zero

But 0 * an undefined limit is still undefined, right?

Last edited: May 26, 2014
2. May 26, 2014

### Staff: Mentor

Wrong :tongue2:. $0 \times \infty$ is undefined, but $0 \times a$ where $|a| < \infty$ is always 0.

3. May 26, 2014

### Waxterzz

but

lim (exp(ik-a)*x) to infinity equals lim exp (ikx) times lim exp(-ax) so it is ∞ * 0 ?

4. May 26, 2014

### pasmith

How do you think $e^{-ax} \cos(kx)$ behaves as $x \to \infty$?

If $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ exist, then it follows that $$\lim_{x \to \infty} f(x)g(x) = \left(\lim_{x \to \infty} f(x) \right)\left(\lim_{x \to \infty} g(x) \right).$$ However, even if one or both of $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ do not exist, it might still be the case that $\lim_{x \to \infty} f(x)g(x)$ exists.

5. May 26, 2014

### Staff: Mentor

$$0 \leq \left| e^{i k x} \right| \leq 1$$

6. May 26, 2014

### Waxterzz

So lim exp (ikx) does not exist , but lim exp( |ikx|) does?

But how make it work in lim (exp(ik-a)x)

Do you know where this rule can be found or what this rule can be called?

But then you've already calcultated lim exp (-ax) that turns zero and lim exp (ikx) does not become a number?

I really don't get it.

If you calculate the limit as a whole (lim exp(ik-a)), you don't get a * 0 or you do?

7. May 26, 2014

### Waxterzz

http://en.wikipedia.org/wiki/Absolute_convergence

So because of that absolute value (the norm?) that is always 1 it converges? I think I got it.

e^ikx converges? |e^ikx| = 1 right ? e^ikx * e^-ikx = e ^ (ikx-ikx) = e^0 = 1 right

but why is the limit not defined?

Wait, because it's periodic. Limit is not defined, but the max and min stays fixed. And thats why when you multiply with a function that goes to zero at infinity the product of the two functions also does?

God, I was too tired to make sense out of anything I think.

8. May 26, 2014

### Staff: Mentor

Sorry, I should've been more clear. Indeed
$$\lim_{x \rightarrow \infty} e^{i k x} = \textrm{undefined}$$
but $0 \leq \left| e^{i k x} \right| \leq 1$, which means it is finite. I use the absolute value since the result is complex, but the important thing is that $e^{i k x}$ stays bounded even as $x \rightarrow \infty$.

Therefore, since
$$\lim_{x \rightarrow \infty} e^{-a x} = 0$$
you have
$$\lim_{x \rightarrow \infty} e^{i k x} e^{-a x} = 0$$
The fact that the limit of the first term is undefined doesn't affect the fact that the product of something finite with zero will always be zero.