Let $p(x)$ be a polynomial and consider the limit $\lim_{h\to0} \frac{p(x+3h)+p(x-3h)-2p(x)}{h^2}$ I am supposed to prove that it is equal to $9p''(x)$ which is simple using l'hopital rules.
However, I am failing when trying to solve it using a substitution and I was wondering what I am doing wrong.
Let $u=3h$ since $h\to0$ so $u\to0$. So we plug it in to get $9\lim_{u\to0}\frac{p(x+u)+p(x-u)-2p(x)}{u^2}=9\lim_{u\to0}\frac{1}{u}(\frac{p(x+u)-p(x)}{u}+\frac{p(x-u)-p(x)}{u})=9\lim_{u\to0}\frac{1}{u}(p'(x)-p'(x))=0$ which is clearly false.
What did I do wrong? My guess is that I divided the limit up to get $p'(x)$ but $\frac{1}{u}$ does not have a limit at $0$ and so I could not do that. However, what can I do to avoid this problem?
Have you tried Taylor polynomials ?
$$P(x+3h)= P(x) + 3hp'(x) + 9h^2P''(x)/2 + 27h^3P'''(x)/6 +....$$
$$P(x-3h)= P(x) - 3hp'(x) + 9h^2P''(x)/2 - 27h^3P'''(x)/6 +....$$
$$\lim_{h\to0} \frac{p(x+3h)+p(x-3h)-2p(x)}{h^2}$$
$$=\lim_{h\to0} \frac{9h^2P''(x)+O(h^4)}{h^2} = 9P''(x)$$