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 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: 20,978 Thanks: 2229 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: 6,823 Thanks: 723 $\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x->0. July 31st, 2017, 12:19 PM   #5
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 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
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 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded 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

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