$\lim\limits_{h\to0}\frac{f(x-h)-f(x)}{h}=-f'(x)$ if $f'(x)$ exist

Question

Suppose for $f:\Bbb R\to \Bbb R$, and $x\in \Bbb R$, $f'(x)$ exists. How to show $\lim\limits_{h\to0}\dfrac{f(x-h)-f(x)}{h}=-f'(x)$?

My try: Since $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=f'(x)$, for every $\varepsilon>0$, there is a $\delta>0$ such that $0<|h|<\delta$ implies $\left|\dfrac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$. Now I want to show for this $\delta$, $\left|\dfrac{f(x-h)-f(x)}{h}+f'(x)\right|=\left|\dfrac{f(x+(-h))-f(x)}{h}-(-f'(x))\right|<\varepsilon$ holds, but since I don't know $-f(x-h)=f(x+h)$, I can't proceed.

Also: Does $\lim\limits_{h\to0}\dfrac{f(x-h)-f(x)}{h}=-f'(x)$ hold for one sided derivatives?

Answer 1

\begin{align*} \left|\dfrac{f(x-h)-f(x)}{h}-(-f'(x))\right|=\left|\dfrac{f(x+(-h))-f(x)}{-h}-f'(x)\right|<\epsilon. \end{align*} The $-h$ in the second one serves as an $h$ in the $\left|\dfrac{f(x+h)-f(x)}{h}-f'(x)\right|<\epsilon$.

Answer 2

Simply note that for $h=-k\to 0$

$$\lim\limits_{h\to0}\dfrac{f(x-h)-f(x)}{h}=\lim\limits_{k\to0}\dfrac{f(x+k)-f(x)}{-k}=-\lim\limits_{k\to0}\dfrac{f(x+k)-f(x)}{k}=-f'(x)$$

Answer 3

Since $h\mapsto -h$ is an invertible continuous map, you have $$ \lim_{h\to0}\frac{f(x-h)-f(x)}{h} = \lim_{h\to0}\frac{f(x+h)-f(x)}{-h} $$ in the sense that one exists if and only if the other one does.