# Solving under the limit sign

1. Jul 30, 2013

### SamRoss

Is it okay to perform operations under the limit sign just as we would if the limit sign were not there? I'm really asking this to better understand how to derive the value of e.

e is defined so that its exponential function is its own derivative. I won't go through all of it (since I'm sure most of the people on this site are familiar) but that amounts to saying that

lim ((e^h-1)/h) = 1
h->0

Now, if we simply had ((e^h-1)/h) = 1 without the limit and wanted to solve for e, we would get

e = (h+1)^(1/h)

Indeed, the true value of e, putting the limit back in, is

e = lim (h+1)^(1/h)
h->0

I have tried this out with a few other functions and it always turns out that performing operations under the limit sign as if it weren't even there and then slapping it back on at the end gives the correct answer, but I just can't convince myself that it's "legal". Can anyone help convince me?

2. Jul 30, 2013

### HallsofIvy

Staff Emeritus
It is as long as you have uniform convergence of the limit. Do you kow what that means?

3. Jul 30, 2013

### SamRoss

I just looked it up and spent about an hour and a half making some sense of it. I'm still not sure I see the relevance, though. Isn't uniform convergence defined for sequences of functions? (e^h-1)/h is just a single function with h as the variable. Now, I suppose we could get around that by trivially multiplying it by n/n and then defining the sequence of functions as

fn(h) = (n/n)((e^h-1)/h)

However, I still don't see how a uniformly convergent sequence of functions allows us to perform normal algebra under the limit sign. Is there a theorem about that?

4. Jul 30, 2013

### hilbert2

For example, $lim_{n\rightarrow \infty} \left( \frac{1}{2} \right)^{1/n}=1$ , but if you try to solve under the limit sign by raising both sides of the equation to n:th power, you'd get $\frac{1}{2}=lim_{n\rightarrow \infty}1^{n}$ , which clearly isn't true. In many cases, informal manipulation under the limit sign works, though.

5. Jul 31, 2013

### SamRoss

HallsofIvy said in his post that manipulation under the equal sign works when there is uniform convergence. Is that true? If so, do you know a proof for it?

6. Jul 31, 2013

### hilbert2

I'm not sure what HallsOfIvy means when talking about uniform convergence in this context. As you said, its a concept related to sequences of functions. When talking about the limit of a single function at a certain point, we have the concept of uniform continuity.