July 31st, 2017, 10:56 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 11:47 AM. 
July 31st, 2017, 11:50 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,568 Thanks: 1483 
Was there any relevant context?

July 31st, 2017, 01:11 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 xaxis 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 01:18 PM. Reason: rogue parenthesis 
July 31st, 2017, 01:18 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,436 Thanks: 562 
$\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x>0.

July 31st, 2017, 01:19 PM  #5 
Newbie Joined: Jul 2017 From: US Posts: 7 Thanks: 0  
August 1st, 2017, 02:29 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,436 Thanks: 562  

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