# Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$

Let $f''$ be continuous on $\mathbb{R}$. Show that

$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$

My workings

$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\lim_{h\to0}\frac{f(x+h)-f(x)-[f(x)-f(x-h)]}{h^2}=\frac{\lim_{h\to0}\frac{f(x+h)-f(x)}{h}-\lim_{h\to0}\frac{f(x)-f(x-h)}{h}}{\lim_{h\to0}h}$$

By the definition of derivative, I move on to the next step. Also, I observe that everything in this question as continuous and differentiable up to $f''(x)$.

$$=\frac{f'(x)-f'(x-h)}{\lim_{h\to0}h}$$

I do not know how to justify the next move but,

$$=\lim_{h\to0}\frac{f'(x)-f'(x-h)}{h}$$

Then by the definition of derivative again,

$$=f''(x-h)$$

Which is so close to the answer. So I shall assume that since $h\to0$ for $x-h$, therefore $x-h=x$? And so,

$$=f''(x)$$

I think i made a crapload of generalization and fallactic errors... I also have another way, which was to work from $f''(x)$ to the LHS. But I realised I assume that the h were the same for $f'(x)$ and $f''(x)$.

Is it normal to be unable to solve this question at the first try? Or am I just too weak in mathematics?

• "Or am I just too weak in mathematics?" — no, never be so hard on yourself. This sort of skill comes with practice. Oct 27, 2012 at 7:06
• And yes, I do suggest trying starting from $f''(x)$, with your manipulations being guided by trying to reach the LHS of your original expression. You're less likely to go astray that way, I think. Oct 27, 2012 at 7:11
• Ok ill try that out and post on here soon :d Oct 27, 2012 at 7:16
• On actually trying this, it seems that (a) it is not at all a straightforward problem, so you shouldn't feel bad about encountering some difficulty, and (b) your approach actually seems a good way. For the last step, you can use the trick of replacing $h$ by $-h$, as the limit does not change. That is, writing $h = -k$, we get $$\lim_{h\to0}\frac{f'(x)-f'(x-h)}{h} = \lim_{-k\to0}\frac{f'(x)-f'(x+k)}{-k} = \lim_{k\to0}\frac{-f'(x)+f'(x+k)}{k} = f''(x)$$. Your overall approach is fine; it's some intermediate steps that have bugs and need to be fixed / made rigorous. Oct 27, 2012 at 11:34
• @SingaporeanDude: See my answer. Do not worry about downvoting! Oct 27, 2012 at 13:57

When you use that $f'(x) = \lim_{h \to 0} \frac {f(x+h)-f(x-h)}{2h}$, you get that $f''(x) = \lim_{k \to 0} \lim_{h \to 0} \frac {f(x+h+k)-f(x+h-k)-f(x-h+k)+f(x-h-k)}{4hk}$.

Let $g_x(h,k)$ be that expression : $f''(x) = \lim_{k \to 0} \lim_{h \to 0} g_x(h,k)$, while what you are given is $\lim_{h \to 0} g_x(h/2,h/2)$.
Both limits are going to $(0,0)$ but not along the same path. So it makes sense that they should be equal under certain conditions, namely if the function $g_x$ can be continuously extended at the point $(h=0,k=0)$ and on the two axis $h=0$, $k=0$: then no matter what path you take in your limit to $(0,0)$, you will get the same result.

In order to show that it is the case, you need to use the fact that $f''$ is continuous. Apply the mean value theorem to the functions $g_{x,h} : k \mapsto f(x+h+k) - f(x-h+k)$ : forall $h,k$, there is a $k'$ such that $|k'|\le |k|$ and $g_{x,h}(k) - g_{x,h}(-k) = 2kg'_{x,h}(k')$, which means that $g_x(h,k)$ simplifies to $\frac{g'_{x,h}(k')}{2h} = \frac{f'(x+h+k') - f'(x-h+k')}{2h}$.
Note that with the continuity of $f'$, this implies that for $h \neq 0$, $g_x$ can be continuously extended at $g_x(h,0)$ by $g_x(h,0) = \frac {f'(x+h)-f'(x-h)}{2h}$ (and similarly on the other axis)

Next we can apply the mean value theorem again, to all the functions $h \mapsto f'(x+h+k')$:
forall $h,k$ there are some $h',k'$ such that $|h'| \le |h|, |k'| \le |k|$, and $g_x(h,k) = f''(x+h'+k')$.

Then, we use the continuity of $f''$ to conclude that $\lim_{(h,k) \to (0,0)} g_x(h,k) = \lim_{(h,k) \to (0,0)} f''(x+h'+k') = f''(x)$

Try applying L'Hospital's Rule to $h$, that is, differentiate with respect to $h$.

Apply Taylor's formula in the form $f(x+h) = f(x) + h f'(x) + h^2 f''(x)/2 + o(h^2 f''(x))$.

• hey can you explain to me why you wrote $f(x+h) = f(x) + h f'(x) + h^2 f''(x)/2 + o(h^2 f''(x))$ instead of $f(x+h) = f(x) + x f'(x) + x^2 f''(x)/2 + o(x^2 f''(x))$ what is the reason you replaced x with h? Jan 23, 2018 at 20:27

Your problem hinges on L'Hospital's Rule for $h$ and the following formula for derivative which known as the symmetric formula

$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h} \,,$$

which implies

$$f''(x) = \lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}\,.$$

See here for details of deriving the above formula.

• -1. Neither $f'(x+h)$ nor $f'(x-h)$ appear in the formula the OP wants to be explained.
– Did
Oct 27, 2012 at 13:10
• @did:Yes it did. Oct 27, 2012 at 13:19
• See here. You know you do not care about the reason.
– Did
Oct 27, 2012 at 13:42
• Ahhh yes, I had forgotten this one: your famous try to work it out. Once again, the problem is not that people would understand incorrectly your so-called solutions (they understand all right), nor that they would be unable to reach a solution themselves (they are able to), nor even that your so-called solutions often omit the crucial step of the proof, or that they are deeply wrong (which can happen to anybody), it is that you refuse to listen to people who explain to you why.
– Did
Oct 27, 2012 at 14:06
• heh its ok guys! I understand it now. Although the explanation was albeit short, and I had to infer a bunch of stuff, but I got it in the end, and it was a fine learning process. cheers to all. Oct 27, 2012 at 14:44

Apply Cauchy mean value theorem for $p(h)=f(x+h)+f(x-h)-2f(x)$, and $g(h)=h^2$, in $[0,h]$, twice.

Since $p(0)=p'(0)=g(0)=g'(0)=0$, you can get

$\frac{p(h)}{h^2}=\frac{p''(z)}{2}=\frac{f''(x+z)+f''(x-z)}{2}\ \text{for z is strictly beteween h and 0}$

Therefore, if you take h goes to zero, then z also goes to zero. Finally, use the contiunuity condition of $f''$.