Solve Derivative Problem: Lim {P(x+3h)+P(x-3h)-2P(x)}/h^2

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The limit problem involves evaluating lim {P(x+3h)+P(x-3h)-2P(x)}/h^2 as h approaches 0, where P is a polynomial function. The suggested answer is D) 9P''(x), which can be confirmed using Taylor series expansions for P(x+3h) and P(x-3h). Higher-order terms become negligible as h approaches 0, simplifying the expression. An alternative method involves applying L'Hopital's rule, differentiating the numerator and denominator twice with respect to h. This approach also leads to the conclusion that the limit equals 9P''(x).
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Hi members,

Could anyone 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
 
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The easiest way is to use Taylor-series expansions of the terms P(x+3h), P(x-3h)
to verify your guess.
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
 
Thank you, arildno

It couldn't be more wonderful solutions.
 
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.
 
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