March 21st, 2019, 03:12 PM  #11  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
$\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_0x \to \infty} \Delta{x} = dx $ The I was told that $\displaystyle \lim_{x_0x \to \infty} \Delta{x} = 0$ 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 reread 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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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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