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?


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

  • $\begingroup$ And divide by two. $\endgroup$ – Rory Daulton Jan 4 '15 at 23:08
  • $\begingroup$ How can you take the minus sign out of the function's input :S how is that true? $\endgroup$ – FigureItOut Jan 4 '15 at 23:18
  • $\begingroup$ Good question: this is the key step. Just replace $h$ with $-h$. Do you see why we can do that, and why it works? $\endgroup$ – user134824 Jan 4 '15 at 23:25
  • $\begingroup$ Yes :) Thanks!!!! $\endgroup$ – FigureItOut Jan 4 '15 at 23:44

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