My Math Forum Indeterminate form limit

 Calculus Calculus Math Forum

 July 31st, 2017, 09:56 AM #1 Newbie   Joined: Jul 2017 From: US Posts: 7 Thanks: 0 Indeterminate form limit Quick question: Wouldn't $\displaystyle \lim_{(x,0)\rightarrow(0,0)}\frac{x^2}{x^4} = \frac{0}{0}$? If so, how did he get the third equation? If we simplify the second equation, we'd have \frac{1}{x^2} which would be undefined, no? Last edited by skipjack; July 31st, 2017 at 10:47 AM.
 July 31st, 2017, 10:50 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,037 Thanks: 1394 Was there any relevant context?
 July 31st, 2017, 12:11 PM #3 Newbie   Joined: Jul 2017 From: US Posts: 7 Thanks: 0 The problem was just a demonstration of how to take limits with multiple variables. By substituting 0 for y, we're testing what the limit would be as we approach from the x-axis so we can see if the limit DNE. Although I believe I understand now where the author was going with this. In the second equation, $\displaystyle lim_{(x,0)\rightarrow(0,0)}\frac{x^2(0)^4}{x^4+3(0 )^4}$ $\displaystyle = lim_{(x,0)\rightarrow(0,0)}\frac{0}{x^4}$ $\displaystyle = 0$ We'd only plug 'n chug the numbers if the function was continuous at 0, but because it isn't, my assumption with the limit being 0/0 arises from false assumption that we would take x=0 and then evaluate $\displaystyle \frac{(0)^2(0)^2}{(0)^4+3(0)^4} = \frac{0}{0}$ Last edited by pst; July 31st, 2017 at 12:18 PM. Reason: rogue parenthesis
 July 31st, 2017, 12:18 PM #4 Global Moderator   Joined: May 2007 Posts: 6,341 Thanks: 532 $\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x->0.
July 31st, 2017, 12:19 PM   #5
Newbie

Joined: Jul 2017
From: US

Posts: 7
Thanks: 0

Quote:
 Originally Posted by mathman $\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x->0.
Right, but the 0 in the numerator would cancel out the x^2 term, right? So we wouldn't be able to take it out of x^4 in the denominator, yes?

August 1st, 2017, 01:29 PM   #6
Global Moderator

Joined: May 2007

Posts: 6,341
Thanks: 532

Quote:
 Originally Posted by pst Right, but the 0 in the numerator would cancel out the x^2 term, right? So we wouldn't be able to take it out of x^4 in the denominator, yes?
We are trying to get a limit as x->0,so setting x=0 defeats the purpose.

 Tags form, indeterminate, limit

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MrDrProfessorPatrick Calculus 7 May 3rd, 2016 01:45 PM MustafaMotani Calculus 2 October 31st, 2014 09:35 AM GeniusBoy Calculus 20 October 5th, 2014 07:05 PM Shamieh Calculus 2 September 4th, 2013 11:24 AM dummiesx3 Calculus 5 July 6th, 2011 03:21 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top