Monday, December 26, 2016

derivatives - 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?


Answer



The factor you're missing is because of the $1/2$ that arises whenever you expand to within second order:



$$f(x) = f(d) + f'(d) (x-d) + 1/2 f''(d) (x-d)^2 + o((x-d)^2).$$




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.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...