If $\exists f''(0)$, show $\lim _{h \rightarrow 0} \frac{f(h)-2f(0)+f(-h)}{h^2}=f''(0)$ Suppose $f$ is differentiable and $\exists f''(0)$.
I want to show
$$\lim _{h \rightarrow 0} \frac{f(h)-2f(0)+f(-h)}{h^2}=f''(0)$$ not using the L'hospital rule.
I know how to calculate with L'hospital rule, but I cannot see how otherwise.
 A: Since $f''(0)$ does exist, $$f(h) = f(0) + h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + o(h^2)$$
$$f(-h) = f(0) - h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + o(h^2)$$
Upon adding we get ,
$$f(h)+f(-h)-2f(0) =  h^2 \cdot f''(0)+o(h^2)$$
$$\frac {f(h)+f(-h)-2f(0)}{h^2} = f''(0) + o(h^2)/h^2$$
Let $h\to 0$ and the desired result is achieved.  
A: Here is one proof which is an adaptation of the proof of Taylor's theorem as given here.

Let $$g(x) =f(x) - \frac{x^2}{2}f''(0)-xf'(0)-f(0)\tag{1}$$ so that $$g(0)=g'(0)=g''(0)=0\tag{2}$$ and then $$g(h) - 2g(0)+g(-h)=f(h)-2f(0)+f(-h)-h^2f''(0)$$ so that our job is done if we show that $$\lim_{h\to 0}\frac{g(h)-2g(0)+g(-h)}{h^2}=0\tag{3}$$ Let's note that by definition of derivative we have $$g'(x) =g'(0)+xg''(0)+o(x)=o(x)$$ and by mean value theorem we have $$g(h) =g(0)+hg'(\xi)=hg'(\xi) $$ for some $\xi$ between $0$ and $h$. And using previous equation we can see that $$g(h) =ho(\xi) =o(h^2)$$ as $0<|\xi|<|h|$. And changing sign of $h$ we get $$g(-h) =o(h^2)$$ Using these we can easily see that $(3)$ holds and we are done. 
A: Use two Taylor series for $f'$ at $0$ using steps $h$ and $-h$.
$$f(h) = f(0) + h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + \frac{h^3 \cdot f'''(0)}{3!} + \frac{h^4 \cdot f''''(0)}{4!} ...$$
$$f(-h) = f(0) - h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} - \frac{h^3 \cdot f'''(0)}{3!} + \frac{h^4 \cdot f''''(0)}{4!} ...$$
Next, lets add them together.
$$f(h) + f(-h) = 2f(0) + 2\frac{h^2 \cdot f''(0)}{2!} + 2\frac{h^4 \cdot f''''(0)}{4!} ...$$
Note the odd powers of $h$ cancel. Solve for $f''(0)$:
$$f''(0) = \frac{f(h) - 2 f(0) + f(-h)}{h^2} + 2\frac{h^2 \cdot f''''(0)}{4!} ...$$
What we have so far is the derivation to the numerical approximation to the second derivative use in computational science. But, just take your  limit and the second and all following terms disappear because h is in the numerator.
