# Given a differentiable function $f$, prove that $\lim_{h\to 0}\frac{f\left(x + \frac{h}{2}\right)-f\left(x - \frac{h}{2}\right)}{h} = f'(x)$.

I'm trying to prove the following statement

Given a differentiable function $$f:\mathbb{R}\to \mathbb{R}$$, prove that $$\lim_{h \to 0} \frac{f\left(x + \frac{h}{2}\right)-f\left(x - \frac{h}{2}\right) }{h} = f'(x)$$

I know $$2$$ definitions for the derivative of a function $$f$$: $$\lim_{h \to 0} \frac{f\left(x +h\right)-f\left(x\right) }{h} \qquad \text{and} \qquad \lim_{a \to x} \frac{f\left(x \right)-f\left(a\right) }{x-a}$$ so my idea was to try to arrange the limit in question into one of these $$2$$ forms.

I proceeded to take $$x^* = x - \frac{h}{2}$$ to do a change of variable, in which case my limit would end up looking like $$\lim_{h \to 0} \frac{f\left(x^* +h\right)-f\left(x^*\right) }{h} = f'(x^*)$$ which would seem to imply that this is equal to $$f'\left(x - \frac{h}{2}\right)\neq f'(x)$$ using the first definition of the derivative. I know that since $$h \to 0$$, then $$f'\left(x - \frac{h}{2}\right)$$ and $$f'(x)$$ are the same thing, but since the paramater $$h$$ is included in the substitution I'm making I don't know how to account for this after I took the limit.

I think my problem is just a basic concept misconception, but I can't seem to find a way to rigorously justify the steps I need to take to transform the limit into one of the definitions that I know. Could anyone tell me what I'm doing wrong or how I could correctly structure this argument to make it rigorous? Thank you!

• $\lim_{h \to 0} f'(x-\frac h2) = f'(x)$ only if $f'$ is continuous at $x.$ – md2perpe Jul 4 at 19:28

$$\lim_{h \to 0} \frac{f\left(x + \dfrac{h}{2}\right)-f\left(x - \dfrac{h}{2}\right) }{h}=\lim_{h \to 0} \dfrac{f\left(x + \dfrac{h}{2}\right)-f(x)+f(x)-f\left(x - \dfrac{h}{2}\right) }{h} \\=\lim_{h \to 0} \dfrac{f\left(x + \dfrac{h}{2}\right)-f(x)}{2\dfrac h2} +\lim_{h \to 0} \dfrac{f\left(x - \dfrac{h}{2}\right)-f(x) }{-2\dfrac h2}=2\frac{f'(x)}2.$$
Note that the reciprocal is not true, because $$f(x)$$ does not appear in the symmetric formula. So the equivalence is one-way.
• Thank you so much for your answer! I have a question about the last remark. To make the reciprocal true, would it then suffice to ask that $f(x)$ be defined for all $x \in \mathbb{R}$? – Robert Lee Jul 4 at 19:54
• @RobertLee: not at all. It needs to be continuous at $x$. – Yves Daoust Jul 4 at 19:58
• Does the hypothesis that $f$ is differentiable on $\mathbb{R}$ not imply that it's also continous over the reals? – Robert Lee Jul 5 at 0:32
• For a classic example, the limit exists for $f(x) = |x|$ at $x = 0$, but clearly $f'(0)$ does not exist. – Charles Hudgins Jul 5 at 4:36
\begin{align} & \frac{f\left(x + \frac{h}{2}\right)-f\left(x - \frac{h}{2}\right) }{h} \\[12pt] = {} & \frac 1 2\left( \frac{f\left(x + \frac{h}{2}\right) - f(x)}{h/2} + \frac{f(x) -f\left(x - \frac{h}{2}\right)}{h/2} \right) \\[8pt] \to {} & \frac 1 2\Big( f'(x) + f'(x)\Big) \quad \text{as } h\to0. \end{align}