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March 21st, 2019, 03:12 PM   #11
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Quote:
 Originally Posted by NineDivines The most embarrassing thing about this post is not on the OP but the glorified “Math Team” who don’t give a sh*t. Who can spot what’s wrong with the OP’s definition of derivative? How on Earth that slipped past by this so called “team” I don’t know. Either give up your gloves for the sake of your reputation or be active to correct people.
My thinking was that:

$\displaystyle \frac{\Delta{y}}{\Delta{x}} = \frac{f(x+h) - f(x)}{h}$

So if I multiply both sides with lim:
$\displaystyle \frac{\lim_{h \to \infty}\Delta{y}}{\lim_{h \to \infty} \Delta{x}} = \lim_{h \to \infty} \frac{f(x+h) - f(x)}{h}$

And this is why I assumed that:
$\displaystyle \lim_{h \to \infty} \Delta{x} = dx \Leftrightarrow \lim_{x_0-x \to \infty} \Delta{x} = dx$

The I was told that $\displaystyle \lim_{x_0-x \to \infty} \Delta{x} = 0$ which is true. They also told me that $\displaystyle \frac{d}{dx}$ is just an operator (symbol).

I got the idea now not entirely but I will try to understand better next time I'll re-read calculus, having in mind what I was told here.

Last edited by babaliaris; March 21st, 2019 at 03:16 PM.

March 21st, 2019, 04:51 PM   #12
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Quote:
 Originally Posted by babaliaris So if I multiply both sides with lim: $\displaystyle \frac{\lim_{h \to \infty}\Delta{y}}{\lim_{h \to \infty} \Delta{x}} = \lim_{h \to \infty} \frac{f(x+h) - f(x)}{h}$

Your limit is as h goes to 0, not as h goes to infinity.

You are not "multiplying both sides with lim." lim is an operator. It acts on both sides of the equation but it is NOT a number!

Yes, it is true that $\displaystyle \lim_{h \to 0} \dfrac{\Delta y}{ \Delta x} \to \dfrac{dy}{dx}$ but bear in mind that this is not an equality. The limit of $\displaystyle \dfrac{ \Delta y}{ \Delta x}$ as h goes to 0 is $\displaystyle \dfrac{dy}{dx}$ is what is true.

It sounds like a minor point of language but I assure you it is critical in your thinking. There is no value h such that $\displaystyle \dfrac{ \Delta y}{ \Delta x} = \dfrac{dy}{dx}$

-Dan

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