A limit with an intuitive and wrong answer In my last question I asked about a limit used in my exploration of tangent circles and whatnot. 
I decided to come up with a more direct approach to my problem, and now I only have to evaluate the limit 
$$ \lim_{d\to x} \frac{\dfrac{f(x)-f(d)}{x-d}-f'(d)}{x-d}$$
Intuition would yield the answer is the second derivative of $f$ at $x$. However, by expanding and whatnot and then using l'Hôpital, as well as plugging in some sample $f$'s, I arrive at half of the second derivative. Why is my intuition wrong? 
 A: The factor you're missing is because of the $1/2$ that arises whenever you expand to within second order:
$$f(x) = f(d) + f'(d) (x-d) + 1/2 f''(d) (x-d)^2 + o((x-d)^2).$$
Substitute and you have
$$\frac{f'(d) + 1/2 f''(d) (x-d) + o(x-d) - f'(d)}{x-d} = 1/2 f''(d) + o(1).$$
This should explain your results. egreg's answer shows a way to see this using only L'Hopital's rule; here the $2$ arises from the derivative of $(x-d)^2$ in the denominator.
Here is some clarification of notation, in case you're not familiar with it already. In the above, $o(g(x))$ is a replacement for a function, which I am not specifying, and which has the property $\lim_{x \to d} \frac{o(g(x))}{g(x)} = 0$. This is called "little oh notation", and it is fairly standard.
A: Let's assume $f$ is differentiable in a neighborhood of $x$ and the second derivative is continuous at $x$; then we can apply l'Hôpital's theorem to the limit in the form
\begin{align}
\lim_{d\to x}\frac{f(x)-f(d)-(x-d)f'(d)}{(x-d)^2}
&\overset{\mathrm{H}}{=}
\lim_{d\to x}\frac{-f'(d)+f'(d)-(x-d)f''(d)}{-2(x-d)}\\
&=\lim_{d\to x}\frac{f''(d)}{2}\\
&=\frac{f''(d)}{2}
\end{align}
Think to Taylor's expansion. Of course these are not the minimal hypotheses for the result to hold.
A: The other answers are great. I just want to zero in on the crucial difference between what you got and what you expected. If you wrote
$$\lim_{d\to x} \frac{f'(x)-f'(d)}{x-d}$$
that would be $f''(x)$. Note that the quantity $f'(x)$ appearing the numerator is the value of $f'$ at $x$. However, in your numerator, you have $\frac{f(x)-f(d)}{x-d}$. This quantity is the average value of $f'$ over the interval $[x,d]$.
