# 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.

• What is your definition of the derivative? – Ahmed S. Attaalla Sep 2 '18 at 18:10
• Let $x_1 = x_0 + \Delta x$. Then let $2 \Delta x = c$. The rest follows by simple logic. – Drew Christensen Sep 2 '18 at 18:31

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}

• Quick clarification: What did you do between steps 3 to 4? – S. Sharma Sep 2 '18 at 19:30
• I've replaced $\Delta x$ with $-\Delta x$ – Elliot G Sep 2 '18 at 19:56

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.

• +1 for "can't stand $\Delta x$". – Paramanand Singh Sep 3 '18 at 3:57