Evaluate a twice differentiable limit

Question

Evaluate $$\lim_{h \rightarrow 0} \frac{f(x+h) -2f(x) + f(x-h) } { h^2}$$

if $f$ is a twice differentiable function.

I'm not sure how to understand this problem. If I differentiate the numerator I get $f'(x+h) - 2f'(x) + f'(x-h)$ but that doesn't seem to take me anywhere?

regroup you get $f'(x+h)-f'(x)-(f'(x)-f'(x-h))$. – TTY Mar 15 '14 at 15:18 — TTY, Mar 15 '14 at 15:18
A related problem. – Mhenni Benghorbal Mar 15 '14 at 18:08 — Mhenni Benghorbal, Mar 15 '14 at 18:08

score 3 · Answer 1 · answered Mar 15 '14 at 15:21

3

Hint

Use the Taylor formula: $$f(x\pm h)=f(x)\pm hf'(x)+\frac{h^2}2f''(x)+o(h^2)$$

answered Mar 15 '14 at 15:21

ketan · Accepted Answer · 2014-03-15T15:30:24.670

0

we have $$\lim_{h \rightarrow 0} \frac{f(x+h) -2f(x) + f(x-h) } { h^2}$$ apply L'hopital, differentiate wrt h , u get $$\frac{f'(x+h) - f'(x-h) } { 2h}$$ again by L'hopital u get

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

edited Mar 15 '14 at 15:30

answered Mar 15 '14 at 15:22

ketan

2,213

Why does $-2f(x)$ disappear when you differentiate? – Paze Mar 15 '14 at 15:30
we are to differentiate w.r.t h , so 2f(x) is constant – ketan Mar 15 '14 at 15:31
Ah, I see. Thanks. – Paze Mar 15 '14 at 15:33

score 0 · Answer 3 · answered Mar 15 '14 at 15:25

0

Hint

Rewrite

$$ [f(x+h) -2f(x) + f(x-h)] =[f(x+h)-f(x)] -[f(x)-f(x-h)]$$

Divide each portion by $h$. What does that become ? Do it again to arrive to ...

I am sure that you can take from here.

answered Mar 15 '14 at 15:25

Claude Leibovici

260,315

Evaluate a twice differentiable limit

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