My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree5Thanks
LinkBack Thread Tools Display Modes
March 21st, 2019, 04:12 PM   #11
Senior Member
Joined: Oct 2015
From: Greece

Posts: 139
Thanks: 8

Originally Posted by NineDivines View Post
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:

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

So if I multiply both sides with lim:
\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:
\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 04:16 PM.
babaliaris is offline  
March 21st, 2019, 05:51 PM   #12
Math Team
topsquark's Avatar
Joined: May 2013
From: The Astral plane

Posts: 2,340
Thanks: 983

Math Focus: Wibbly wobbly timey-wimey stuff.
Originally Posted by babaliaris View Post
So if I multiply both sides with lim:
\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 $\displaystyle \dfrac{ \Delta y}{ \Delta x} = \dfrac{dy}{dx}$

Thanks from babaliaris
topsquark is offline  

  My Math Forum > College Math Forum > Calculus

Δx, limit

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
upper limit = lower limit implies convergence zylo Calculus 13 May 31st, 2017 01:53 PM
Limit yo79 Math Events 1 February 16th, 2014 08:28 AM
Limit 2 yo79 Math Events 3 November 23rd, 2013 09:42 AM
Limit Superior and Limit Inferior veronicak5678 Real Analysis 4 August 22nd, 2011 11:07 AM
when should we evaluate left limit and right limit? conjecture Calculus 1 July 24th, 2008 02:14 PM

Copyright © 2019 My Math Forum. All rights reserved.