Timeline for Why do we teach calculus students the derivative as a limit?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2020 at 17:31 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 18 characters in body
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| Sep 8, 2020 at 17:29 | comment | added | LSpice | I read your 'if' statement as an equivalence, which was why I objected; and so I was wrong because I misread but also, as you point out, because my proposed correction doesn't work. Thank you for the clarification (of my misreading) and for your counterexample. | |
| Sep 8, 2020 at 16:22 | comment | added | Michael Hardy | @LSpice : You're quite mistaken. If $a=0$ and $f(x) = x^3,$ then the line $y=0$ goes from above the curve to the left of $0$ and below to the right, but the slope of that line is not less than $f'(a).$ And nothing in the first definition above implies $x^2\sin(1/x)$ is always positive on one side of $0$ and negative on the other side. Rather than definition says that for every line through $(0,0)$ with positive slope, there is an interval about $0$ for which the line lies above $x^2\sin(1/x)$ to the right of $0$ and below to the left, and similarly for lines with negative slope. $\qquad$ | |
| Sep 8, 2020 at 12:10 | comment | added | LSpice | I think that the first definition should be characterising when $f'(a) > m$, not when $f'(a) \ge m$. Otherwise, with $m = 0$, it incorrectly tells us that there is an interval around $x = 0$ where $f(x) = x^2\sin(1/x)$ is always greater than $0$ on one side, and always less than $0$ on the other side. | |
| S Mar 3, 2015 at 0:36 | history | answered | Michael Hardy | CC BY-SA 3.0 | |
| S Mar 3, 2015 at 0:36 | history | made wiki | Post Made Community Wiki by Michael Hardy |