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
limd→xf(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?
Answer
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)+1/2f″
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.
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