Find $\lim_\limits{h\to 0}{\frac{f(a+h)-f(a-h)}{h}}$ 
Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that:
  $$\lim_\limits{x\to a}{\frac{f(x)-f(a)}{x-a}}=2$$
Find, if it exists, the $\lim_\limits{h\to 0}{\frac{f(a+h)-f(a-h)}{h}}$ without using derivatives and integrals.

So, I have tried the following:
$$\lim_\limits{x\to a}{\frac{f(x)-f(a)}{x-a}}=2 \Leftrightarrow \lim_\limits{h\to 0}{\frac{f(a+h)-f(a)}{h}}=2$$
Any hint how to continue?
 A: Hint. You may write, as $h \to 0$,
$$
\frac{f(a+h)-f(a-h)}{h}=\frac{f(a+h)-f(a)}{h}-\frac{f(a-h)-f(a)}{h}.
$$
A: You can use the same idea.
$$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\rightarrow 0}\frac{f(a)-f(a-h)}{h}=2.$$
So you can sum this limits
$$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}+\lim_{h\rightarrow 0}\frac{f(a)-f(a-h)}{h}=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a-h)}{h}=2+2=4.$$
A: We have an existing limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=2$$ $$\implies LHL=\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}=2$$ Let, $x=a-h \implies h\to 0\ as\ x\to a$ $$\implies LHL=\lim_{h\to 0}\frac{f(a-h)-f(a)}{(a-h)-a}=2$$ $$\implies \lim_{h\to 0}\frac{f(a-h)-f(a)}{-h}=2$$ $$\implies \lim_{h\to 0}\frac{f(a-h)-f(a)}{h}=-2\tag 1$$
Similarly, we have $$ RHL=\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}=2$$ Let, $x=a+h \implies h\to 0\ as\ x\to a$ $$\implies RHL=\lim_{h\to 0}\frac{f(a+h)-f(a)}{(a+h)-a}=2$$ $$\implies \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=2$$ $$\implies \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=2\tag 2$$ Now, subtracting (1) from (2), we get $$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}-\lim_{h\to 0}\frac{f(a-h)-f(a)}{h}=2-(-2)$$ $$\implies \lim_{h\to 0}\frac{f(a+h)-f(a)-f(a-h)+f(a)}{h}=4$$ $$\implies \color{blue}{\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{h}=4}$$
A: $$\lim_\limits{h\to a}{\frac{f(x)-f(a)}{x-a}}=2 \Leftrightarrow \lim_\limits{h\to 0}{\frac{f(a+h)-f(a)}{h}}=2 \rightarrow f'(a)=2\\$$ now  add $\pm f(a)$ 
$$\lim_\limits{h\to 0}{\frac{f(a+h)-f(a-h)}{h}}=\\\lim_\limits{h\to 0}{\frac{f(a+h)-f(a) +f(a)-f(a-h)}{h}}=\\\lim_\limits{h\to 0}{\frac{f(a+h)-f(a)}{h}}+\lim_\limits{h\to 0}{\frac{f(a)-f(a-h)}{h}}=\\2+2$$ for the second limit $-h=H \rightarrow 0 \\\lim_\limits{h\to 0}{\frac{f(a)-f(a-h)}{h}}=\\\lim_\limits{H\to 0}{\frac{f(a)-f(a+H)}{-H}}=\\\lim_\limits{H\to 0}{\frac{f(a+H)-f(a)}{H}}=2 $
