Limit of a function tending to zero.

Question

If $F(t)$ is twice differentiable at $x$ and $$G(h)=\max_{t\in(0,h)}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right],$$ where $x$ is fixed; then how can we show that $\displaystyle\lim_{h\to 0}G(h)=0$.

There are abstract duplicates of this question here and here. — Zev Chonoles, Jun 09 '12 at 15:48

André Nicolas · Accepted Answer · 2012-06-09T16:02:20.047

1

Hint: $$\frac{F'(x+t)-F'(x-t)}{t}=\frac{F'(x+t)-F'(x)}{t}+\frac{F'(x-t)-F'(x)}{-t}.$$

edited Jun 09 '12 at 16:02

answered Jun 09 '12 at 15:46

André Nicolas

507,029

@Thomas: Indeed very useful! I started with the obvious one, then tried to give a little more. – André Nicolas Jun 09 '12 at 15:57
You removed the $t$, I removed my comment. A pity, this equation would have allowed me to show (no, I won't tell ;-) – Jun 09 '12 at 15:59
@Kns: I was being too indirect. Have changed the hint, added primes. Note that each part is related to the derivative of $F'$. – André Nicolas Jun 09 '12 at 16:05

haohaolee · Answer 2 · 2012-06-09T17:19:34.077

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Since $h\to0$ means $t\to0$, so $$\displaystyle\lim_{h\to 0}G(h) = \lim_{t\to 0}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right]$$ apply l'Hôpital's rule，we can get $$ \lim_{t\to 0}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right] = \lim_{t \to 0}\left[\frac{F''(x+t)+F''(x-t)}{2}-F''(x)\right] = 0$$

ps: your tags include real analysis, I assume $x\in \mathbf{R^{n}}$, although real analysis isn't only about real numbers. I dou't know if l'Hôpital's rule can apply under other situation.

edited Jun 09 '12 at 17:19

answered Jun 09 '12 at 16:04

haohaolee

403

(This was my first intent, too: L'Hospital) I've a doubt here: in the RHS the $,t's,$ disappeared while you still are taking the limit when $,t\to 0,$ , but this is equivalent to plug $,t=0,$ after applying L'Hospital, which is justified if $,F'',$ is continuous in $,x,$, something we can't know... – DonAntonio Jun 09 '12 at 16:27
If I recall correctly, L'Hospital does not require the derivative to be continuous(namely $F''$ here), the existent of the derivative in the area is enough – haohaolee Jun 09 '12 at 17:05
Indeed it doesn't, but you require it to put $$\lim_{t\to 0}\frac{F''(x+t)+F''(x-t)}{2}=\frac{F''(x)+F''(x)}{2}$$which is what in fact you did, or, of course, justify otherwise this equality. – DonAntonio Jun 09 '12 at 17:13
oh, you are right, my fault, thanks. edit it. And then I get what you meant.... thinking now – haohaolee Jun 09 '12 at 17:18
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seems it is not appropriate here. Thanks for the reminding – haohaolee Jun 09 '12 at 17:28

score 1 · Answer 3 · edited Apr 13 '17 at 12:21

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Hint: By definition

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

Couple this with Andre's comment

edited Apr 13 '17 at 12:21

Community

1

answered Jun 09 '12 at 16:07

Eelvex

1,414

score 1 · Answer 4 · answered Jun 09 '12 at 16:09

Since $\,t\in(,h)\,$ , we have that $\,h\to 0\Longrightarrow t\to 0\,$ , so: $$\lim_{t\to 0}\frac{F'(x+t)-F'(x-t)}{2t}=\lim_{t\to 0}\frac{1}{2}\left[\frac{F'(x+t)-F'(x)}{t}+\frac{F'(x-t)-F'(x)}{-t}\right]$$ and you get what you want since we know $\,F''(x)\,$ exists, so the limit defining this second derivative exists.

Limit of a function tending to zero.

4 Answers4