# Proof Help: Proving that limit exists and equals the derivative

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.

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

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!!

• Your EDIT is fine. However, you should have in mind that it is true only because both $\frac{1}{2}\lim_{h→0}\frac{f(x + h)-f(x)}{h}$ and $\frac{1}{2}\lim_{h→0}\frac{f(x )-f(x-h)}{h}$ exist. I mean: the limits of a sum may exist, while the limit of both term do not. May 16 '18 at 19:55
• The question says that the function is differentiable on its domain. Can that be used to justify that?
– Alea
May 16 '18 at 20:00
• What you wrote is fine even if you suppose that $f$ is differentiable at $x$ only. I'm just warning on the fact that the existence of the limit of a sum do not imply the existence of the limits of both terms. May 16 '18 at 20:06
• The first line of your proof seems incorrect. That is not the definition of $f'(x).$
– zhw.
May 16 '18 at 20:46
• @A.Asad We can write simply $$\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}=\frac12\lim_{h→0}\frac{f(x + h) − f(x))}{h}-\frac12\lim_{h→0}\frac{f(x - h) − f(x)}{-h}$$
– user
May 17 '18 at 6:44

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.

• Where did you get the two results in the bullet points from? I'm afraid I haven't seen them (or at least I can't recall). Do they come from a particular theorem or or result?
– Alea
May 16 '18 at 19:53
• @A.Asad For functions of one variable the concept is completely equivalent to that of derivative, refer to en.wikipedia.org/wiki/Differential_of_a_function#Definition . For functions of several variables the two concept are different and in some sense differential is the most important one.
– user
May 16 '18 at 20:03
• @A.Asad Note that the little-o notation indicates something which goes to zero with an order $h^n$ with $n>1$, thus $$f(x+h)=f(x)+f'(x)h+o(h)\iff f'(x)=\frac{f(x+h)-f(x)}{h}+\frac{o(h)}{h}$$ and taking the limit since $\frac{o(h)}{h}\to 0$ we obtain the equivalent definition of derivative.
– user
May 16 '18 at 20:06

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.

• Downvoting. The question asked to "verify what [the questioner has] done"; you have given a completely different proof. May 16 '18 at 19:32
• Thank you. Edit: pointed out what the questioner lacks of. May 16 '18 at 19:38
• Thank you for the edit. Removing the downvote and upvoting. May 16 '18 at 19:40
• I did that. However, the problem is that I should be getting $h$ in the numerator which I'm not. Can I take the $1/2$ outside, compute the two limits, giving me $\frac{1}{2}f'(x)$ which added together would give me $f(x)$?
– Alea
May 16 '18 at 19:44
• So the question you should focus on thinking is: can I pull a constant out of a limit? May 16 '18 at 19:46

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 $\frac{|h|-|-h|}{2h}$

• I'm not quite sure how to incorporate that into the numerator. I was thinking of adding and subtracting $f(x)$ to the expression in the numerator but that wasn't leading me anywhere.
– Alea
May 16 '18 at 19:38
• @A.Asad I think that's a good idea. The next step would be to separate that into two derivative-like expressions. May 16 '18 at 19:39
• Can you please check my edit?
– Alea
May 16 '18 at 19:51
• @A.Asad Looks very good! The only thing I'd be a little more clear about is your claim that $\lim_{h\to0}\frac{f(x)-f(x-h)}{h}=f'(x)$. It is true, but it's not quite the definition of the derivative, so perhaps you could justify it. The "limit breaking" is allowed according to the limit law that the limit of a sum is the sum of the limits. May 16 '18 at 19:54
• Thank you! And, for the justification, I could write $f(x) = f((x-h)+h)$ so it is in the form as that required by the definition?
– Alea
May 16 '18 at 19:58

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 $\lim\limits_{h \to 0}\epsilon(h) = 0$.