# if $f(x)$ is differentiable at a x, prove that: $\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$

If $f(x)$ is differentiable at x, I need to prove that $\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$ exist and is finite.

so if $f(x)$ is differentiable at a $x$, the difference quotient exist for this point, and also $f(x)$ must be continuous at x as well

so that mean that: $\lim_{h\to0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}$

and that: $\lim_{x\to x_0^+}=\lim_{x\to x_0^-}=f(x_0)$

I know I should probably use arithmetic limit laws to prove this but I can't see how what I figured out could help me. any help with that?

Thanks!

• Use $f(x + h) - f(x - h) = (f(x + h) - f(x)) + (f(x) - f(x - h))$. – user203787 Jan 4 '15 at 22:39
• Still, how do I continue for $\frac{-f(x-h)+f(x)}{2h}$? – FigureItOut Jan 4 '15 at 22:50
• Each one of $(f(x + h) - f(x))/h , (f(x) - f(x-h))/h \to f'(x)$ as $h \to 0$. – user203787 Jan 4 '15 at 22:53

What does it mean that $f$ is differentiable at $x$? It means that $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=f'(x).$$ This is the same (why?) as $$\lim_{h\to0}\frac{f(x)-f(x-h)}{h}=f'(x).$$ Now add the equations together.
• Good question: this is the key step. Just replace $h$ with $-h$. Do you see why we can do that, and why it works? – user134824 Jan 4 '15 at 23:25