11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The o notation makes sense. thanks! $\endgroup$ – Faraz Masroor Dec 16 '14 at 0:06
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $[x,d]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.