2
$\begingroup$

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!

$\endgroup$
  • 7
    $\begingroup$ 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
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.