# Proving a limit exists

#### DkayD

I've just arrived at university (finally!), and I've been going over some of my A-level calculus work, to make sure I really, really understand it. And mostly I do. But there is one thing that is bugging me. When you attempt to find the derivative of $$a^x$$, you end up with:
$$\frac{\mathrm{d}\! y}{\mathrm{d}\! x}&=&a^{x}\left(\lim_{\delta\! x\rightarrow0}\frac{a^{\delta\! x}-1}{\delta\! x}\right)$$

You then define e to be the number such that:

$$\lim_{x\rightarrow0}\frac{e^{x}-1}{x}=1$$

I'm perfectly comfortable with using that to define a number, I get that we don't need to "see" what the number is for it to be real and useful. What has been really, really bugging me, is how we know that this limit exists in the first place. I tried looking at the $$(\epsilon,\delta)$$definition of a limit, but all I could find were proofs for showing that a certain numerical value of a limit was correct, I could not find anywhere how we prove that a limit exists.

Also, once we know the limit exists, defining a function $$f(x)=\lim_{k\rightarrow0}\frac{x^{k}-1}{k}$$, how do we show that there is an x such that $$f(x)=1$$? I assume that we would use the intermediate value theorem? But of course that assumes that the function is continuous... Which I know it is, because it's $$\ln(x)$$. But how do I show that?

#### Petr Mugver

The simple fact that you wrote ax means that you already have defined it. There are many (equivalent, of course) ways to define the exponential function. Which one do you use?

#### DkayD

The simple fact that you wrote ax means that you already have defined it. There are many (equivalent, of course) ways to define the exponential function. Which one do you use?
It seems that I was taught this back-to-front. The exponential function was never really "defined". It never occurred to me that this may be an issue. I feel stupid now.

Basically, we started by defining it for positive integral powers as repeated addition, and then extended it to negative integers, and then rationals by preserving the two "laws" that held for positive integral powers ($$a^{n}\cdot a^{m}=a^{n+m},\:\left(a^{n}\right)^{m}=a^{nm}$$). We never really touched on how this would be extended to the reals. And yet we were shown a graph of it, and the derivative was "defined" as I said.

Thank you for your help so far, would you mind explaining this, and how it defines the limit?

EDIT: I found a nice, simple definition of ex: $$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$. I have proven (well, I don't know how rigorous it is, but I have proven it to myself) that it is its own derivative given that definition. So it would seem that that limit my teacher used is completely irrelevant and a massive red herring. I feel cheated.

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#### Petr Mugver

Your definition is a good one, but I prefer saying simply that the exponential is that function f(x) that is equal to its derivative, f ' = f, and such that f(0) = 1. From this you can derive all the properties of the exponential. But to see that this definition works you need the Cauchy theorem, which is usually studied at the end of the first year (or maybe semester), and that's why in the first pages of calculus they use the argument you wrote.
I found it as well in my first year book, but there it says something like "we now simply state that the limit exists and that the Neper number exists, and we derive the rest. A more rigorous perspective will be given soon.."
Note that to handle properly the series definition you gave, also in this way you need some tools that you probably don't have yet (convergence of power series, etc.)

So if I were you I would wait a couple of months to know the answer, it's not such a long time!

#### DkayD

Yeah, I guess I can wait a few months! Thanks for the help, this is all very interesting.

"Proving a limit exists"

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