Epsilon Delta definition of a Derivative

Question

The derivative at a specific point $c$ is represented as a limit by: $$ f'(c) = \lim_{x\to c} \frac{f(x) - f(c)}{x - c} $$

It's clear to me that the epsilon delta definition of a derivative at a point $c$ would be:

$$ \forall \epsilon > 0 ~\exists \delta > 0 \forall x: \\ 0 < |x-c| < \delta \rightarrow |\frac{f(x)-f(c)}{x-c} - L| < \epsilon $$

What's unclear to me is how to formally represent the derivative as a function of $x$, rather than only at point $c$. Basically, how would we represent this limit formally (the $\Delta x$ is the part tripping me up):

$$ f'(x) = \lim_{\Delta x\to0} \frac{f(x + \Delta x) - f(x)}{\Delta x} $$

Answer 1

You do know the definition of limit, right? So, just apply it. We can argue that if the derivative at $x$ is $L$, then $$\forall \epsilon > 0 ~\exists \delta > 0 \;\forall \Delta x: \\ 0<|\Delta x| < \delta \implies \left|\frac{f(x + \Delta x) - f(x)}{\Delta x}-L\right| < \epsilon$$ Also, the derivative as a function, $f^\prime (x)$ is simply a function which takes a point in, and spits out the derivative of $f$ at that point. So, you can also define definition of a derivative at a point $c$ and collect all those derivatives under a function.

Answer 2

You may write the last formula as :

$$f'(c) = \lim_{\Delta x\to0} \frac{f(c + \Delta x) - f(c)}{c+\Delta x-c}$$

The last formula appears more intuitive when looking at $f'$ as a function and not at $f'(x)$, which is the value of the derivative at $x$. Indeed the limit of the variable does not depend on the point (i.e. we have $\Delta x \to 0$ and not $x \to c$).

Formally, if you had to write an $\varepsilon-\delta$ proof of a formula for a derivative, for example if $f(x)=x^n$, then $f'(x)=nx^{n-1}$, you have to prove it independently for any $x$. So you may start your proof by "let $x \in \mathbb{R}$, then ...". But once this formula has been proven, then you can easily compute the derivative of any polynomial.

Note

I am unsure I have well spotted your misunderstanding so I gave a quite general answer. I hope it helps anyway.