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?

  • $\begingroup$ Can you say more about how you expanded the expression? $\endgroup$ – Tim Raczkowski Dec 15 '14 at 23:31
  • $\begingroup$ Put the numerator under a common denominator then combined to get $\frac{f(x)-f(d)-f'(d)(x-d)}{(x-d)^2}$, then used l'hopital's to get the 2 in the denominator $\endgroup$ – Faraz Masroor Dec 15 '14 at 23:32
  • $\begingroup$ Now I'm also getting negative half of the second derivative when using l'hopitals, but evaluating with f=x^2 only gives 1. What's going on? $\endgroup$ – Faraz Masroor Dec 15 '14 at 23:39
  • $\begingroup$ @FarazMasroor The second derivative of $f(x)=x^2$ is the constant $2$. $\endgroup$ – egreg Dec 16 '14 at 0:03

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.

  • $\begingroup$ The o notation makes sense. thanks! $\endgroup$ – Faraz Masroor Dec 16 '14 at 0:06

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.

  • $\begingroup$ I'm going to have to do this again because when I did it like this I got a negative. $\endgroup$ – Faraz Masroor Dec 16 '14 at 0:05
  • $\begingroup$ @FarazMasroor You probably just differentiated wrongly. Note the minus sign in the denominator, because we're differentiating with respect to $d$. $\endgroup$ – egreg Dec 16 '14 at 0:06
  • $\begingroup$ Yup that's where it was. Thanks. $\endgroup$ – Faraz Masroor Dec 16 '14 at 0:08

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 $[x,d]$.


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