Calculus Proof Regarding Dervivative I'm pretty elementary in calculus so bear with me.
I had this problem in my calculus book that I got stumped on:

Prove $$\lim_{\Delta x\to0}\frac{f(x_0+\Delta x) - f(x_0 - \Delta x)}{2\Delta x} = f'(x_0)$$

Here was the proof that I came up with:

(1) $$\frac{f(x_0+\Delta x) - f(x_0 - \Delta x)}{2\Delta x} = \frac{f(x_0+\Delta x) - f(x_0 - \Delta x)}{(x_0 + \Delta x) - (x_0 - \Delta x)}$$
  (1) is the definition of slope, so $$\lim_{\Delta x\to0}\frac{f(x_0+\Delta x) - f(x_0 - \Delta x)}{(x_0 + \Delta x) - (x_0 - \Delta x)} = f'(x_0)$$ as that is the definition of slope.

I feel as if this makes sense but something seems off with this proof. It just doesn't really fell complete.
If anyone can guide me as to what I should do, then that would be appreciated.
 A: I'm assuming that by definition
$$f'(x_0)=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.$$
The key here is that, by definition, if the derivative exists, there is a two-sided limit of the difference quotient. In general, whenever $f$ has a two-sided limit at $c$, we have $\lim\limits_{c\to 0}f(x+c)=\lim\limits_{x\to 0}f(x-c)$. You can think of this as approaching $c$ from the left versus from the right. In reality though, when we write $\lim\limits_{c\to 0}$, we just want to see what happens for small values of $c$, positive or negative.
Here we have
\begin{align}
\lim_{\Delta x\to0}\frac{f(x_0+\Delta x) - f(x_0 - \Delta x)}{2\Delta x} &=\lim_{\Delta x\to 0}\left(\frac{f(x_0+\Delta x)-f(x_0)}{2\Delta x}+\frac{f(x_0)-f(x_0-\Delta x)}{2\Delta x}\right)\\
&=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{2\Delta x}+\lim_{\Delta x\to 0}\frac{f(x_0)-f(x_0-\Delta x)}{2\Delta x}\\
&=\frac 12 f'(x_0)-\lim_{\Delta x\to 0}\frac{f(x_0-\Delta x)-f(x_0)}{2\Delta x}\\
&=\frac 12 f'(x_0)-\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{-2\Delta x}\\
&=\frac 12 f'(x_0)+\frac12 f'(x_0)\\
&=f'(x_0).
\end{align}
A: Apologies, but I can't stand $\Delta x$, so I'll use $h$. Your proof is not correct, because it glosses over the main point.

Saying that $f$ is differentiable at $x$ is equivalent to saying that there exist a number $f'(x)$ and a function $\varphi$ defined over $(-\delta,\delta)$ (for some $\delta>0$) such that
$$
f(x+h)=f(x)+hf'(x)+h\varphi(h)
$$
for every $h\in(-\delta,\delta)$ and
$$
\lim_{h\to0}\varphi(h)=0
$$
Applying it to your situation, for $h\in(-\delta,\delta)$, $h\ne0$,
\begin{align}
\frac{f(x+h)-f(x-h)}{2h}
&=\frac{(f(x)+hf'(x)+h\varphi(h))-(f(x)-hf'(x)-h\varphi(-h))}{2h} \\[6px]
&=f'(x)+\frac{1}{2}\varphi(h)+\frac{1}{2}\varphi(-h)
\end{align}
and therefore
$$
\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}
=
\lim_{h\to0}\left(f'(x)+\frac{1}{2}\varphi(h)+\frac{1}{2}\varphi(-h)\right)=f'(x)
$$
Let's prove the statement on which the proof above is based.
Suppose $f$ is differentiable at $x$ (and defined in a neighborhood of $x$). Then we can define
$$
\varphi(h)=\frac{f(x+h)-f(x)}{h}-f'(x)
$$
for $h\ne0$ and $\varphi(0)=0$, over some interval $(-\delta,\delta)$ such that, for $h\in(-\delta,\delta)$, $f(x+h)$ is defined. Then, by differentiability,
$$
\lim_{h\to0}\varphi(h)=0
$$
Conversely, if the number $f'(x)$ and the function $\varphi$ exist, we have
$$
\frac{f(x+h)-f(x)}{h}=f'(x)+\varphi(h)
$$
and therefore
$$
\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=f'(x)
$$
so $f$ is differentiable at $x$ with derivative the given number $f'(x)$.
Important note. The statement in the question should assume that $f$ is differentiable at the given point.
