What does $\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}}$ evaluate to? Problem:

Suppose that $f$ is twice continuously differentiable within a
neighborhood of the point $a$ and $f''(a) = 4$. Compute \begin{align*}
 \lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} \end{align*}

I have done this work
\begin{align*}
\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} &= \lim_{x \to a} \frac{\frac{f(x)-f(a)}{x-a} - f'(a)}{x-a} \\
&= \lim_{x\to a}\frac{f'(a)-f'(a)}{x-a} \\
&= 0
\end{align*}
This makes sense, but I am not confident in my answer because I didn't use the given that $f''(a) = 4$. Any help would be greatly appreciated.

According to Taylor's theorem, if $f$ is twice differentiable in $a$ (most versions require continuous differentiability, but it isn't actually necessary), then there exists a remainder function $r(x)$ with
$$\lim_{x\to a}\frac{r(x)}{(x-a)^2}=0$$
such that
$$f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+r(x).$$
With a little bit of algebra we get that
$$\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}=\frac{r(x)}{(x-a)^2}+\frac{1}{2}f''(a).$$
The limit now evaluates to $\frac{1}{2}f''(a)$.
