Proof Help: Proving that limit exists and equals the derivative

Question

Suppose that $f : (a, b) → \mathbb{R}$ is differentiable at $x ∈ (a, b)$. Prove that $\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$ exists and equals $f'(x)$ Give an example of a function where the limit exists but the function is not differentiable.

Proof: $\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$ =$\lim_{h→0}\frac{f((x-h) + 2h) − f(x − h)}{2h}$ $= f'(x)$ by definition which exists as $f:(a,b) \to \mathbb{R}$ is differentiable $x \in (a,b)$.

Can anyone verify what I have done? I think I'm going wrong somewhere. Can anyone please help? Also, I can't think of a function whose limit exists but it is not differentiable. I was thinking of $f(x) =|x|$ but don't know if its correct.

Thank you.

EDIT (After the comments and hints received):

$\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$=$\lim_{h→0}\frac{[f(x + h)-f(x)]+[f(x) − f(x − h)]}{2h}$=$\frac{1}{2}\lim_{h→0}\frac{[f(x + h)-f(x)]+[f(x) − f(x− h)]}{h}$ =$\frac{1}{2}[\lim_{h→0}\frac{f(x + h)-f(x)}{h}$ + $\lim_{h→0}\frac{f(x) − f(x − h)}{h}]=\frac{1}{2}[f'(x)+f'(x)]=f'(x)$

Is this ok? I'm particularly concerned about breaking the limits into two. Is that allowed and can anyone give me a proper theorem or a result/justification that allows me to do that?

Thank you!!

Answer 1

If you want a clearer proof, here is one:

From the definition $f'(x) = lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$, try to prove $f'(x) = lim_{h\to 0} \frac{f(x)-f(x-h)}{h}$. Then use the trick that $f(x+h)-f(x-h) = [f(x+h)-f(x)] + [f(x)-f(x-h)]$.

The whole point is we should prove things based on the definitions. A priori, we only have $f'(x-a)=\lim_{h\to 0} \frac{f(x-a+2h)-f(x-a)}{2h}$ for any fixed $a$. While it is tempting to substitute $h$ into $a$ and takes $h$ to $0$, I would recommend not to "jump" in your early learning of mathematics.

Answer 2

When you say that the expression you have is equal to $f'(x)$ "by definition," you should elaborate. What you have written certainly isn't the usual definition of $f'(x)$. I'm not sure you can make your strategy work, and I think you're going to need a different tack. (Hint: The definition of $f'(x)$ requires there be an $f(x)$ somewhere in the numerator; can you add one in?)

Your $|x|$ example seems like it should work. To convince yourself it's correct, try evaluating $\frac{|h|-|-h|}{2h}$

Answer 3

For the first point, I'm very suspicious of your "by definition proof"...

To be more precise, as $f$ is supposed differentiable at $x$, it exists a map $\epsilon : h \mapsto \epsilon(h)$ with $\lim\limits_{h \to 0} \epsilon(h)=0$ such that $f(x+h) = f(x) + f^\prime(x)h + h\epsilon(h)$. Then:

$$\frac{f(x+h) - f(x-h)}{2h} = \frac{f^\prime(x)h - f^\prime(x)(-h) + \epsilon(h)h - h \epsilon(-h)}{2h}= f^\prime(x) + \frac{\epsilon(h)-\epsilon(-h)}{2},$$

which converges to $f^\prime(x)$ with $h$ as $\lim\limits_{h \to 0}\epsilon(h) = 0$.

Answer 4

Recall that for $f(x)$ differentiable at $x ∈ (a, b)$ we have

• $f(x+h)=f(x)+f'(x)h+o(h)$
• $f(x-h)=f(x)-f'(x)h+o(h)$

then

$$\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}=\lim_{h→0}\frac{2f'(x)h+o(h)}{2h}=\lim_{h→0}\frac{2f'(x)+o(1)}{2}=f'(x)$$

Yes your example for $|x|$ is fine.