If $f''$ exists on an interval centered at $x=a$, then $\lim \limits_{h\to 0} \frac{f(a+h) - 2f(a) +f(a-h)}{h^2} = f''(a)$

Question

If $f''$ exists on an interval centered at $x=a$, then $$\lim \limits_{h\to 0} \frac{f(a+h) - 2f(a) +f(a-h)}{h^2} = f''(a)$$

I've seen a few answers involving Taylor's theorem, and a few that were given hints for L'Hopital's. I do have a question though. If we use L'Hopital's rule, then the first derivative gives us the following :

$$ \frac{f'(a+h) - f'(a-h)}{2h}$$

What I don't understand is why people are saying we can only apply L'Hopital's once here, and say instead to "split the numerator with $f'$". What does this mean, to split the numerator? Why can L'Hopital's be used only once?

Thanks

Answer 1

$\dfrac{f'(a+h) - f'(a-h)}{2h}= \dfrac{1}{2}\left(\dfrac{f'(a+h)-f'(a)}{h}+\dfrac{f'(a+(-h)) - f'(a)}{-h}\right)\to \dfrac{1}{2}\left(f''(a)+f''(a)\right)=f''(a)$ when $h \to 0$.