# A derivative problem

1. Jun 10, 2004

### lhuyvn

Hi members,

Could any one help me with the problem following?

If x is a real number and P is a polynomial function, then

lim {P(x+3h)+P(x-3h)-2P(x)}/h^2
h->0

A)0
B)6P'(x)
C)3P''(x)
D)9P''(x)
E) 00

I guess D should be the answer, I need an explanation.

Thank You

2. Jun 10, 2004

### arildno

The easiest way is to use Taylor-series expansions of the terms P(x+3h), P(x-3h)
Then we have, for example:
P(x+3h)=P(x)+P'(x)3h+1/2P''(x)(3h)^2+O(h^3)

O(h^3) is a higher order term, i.e lim h->0 O(h^3)/h^2=0

3. Jun 10, 2004

### lhuyvn

Thank you, arildno

It couldn't be more wonderful solutions.

4. Nov 12, 2010

### lobsterism

It looks messy, but it's really no different from any other limit problem. What is typically the easiest way to find the limit of 0/0? L'Hopital's rule!

Differentiate top and bottom with respect to h (not x!) twice, using chain rule for top terms. So the first differentiation gives [3P'(x+3h)-3P'(x-3h)]/2h (notice the third term has no h, so drops out). The second round, you get [9P''(x+3h)+9P''(x-3h)]/2. Then setting h=0 gives the desired answer.