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 June 7th, 2011, 05:30 AM #1 Newbie   Joined: Jun 2011 Posts: 2 Thanks: 0 How can I solve this limit problem Hi, How can I solve this problem Limit of (x+y)^2/(x^2+y^2)as (x,y) approaches (0,0)? Thank,
 June 7th, 2011, 05:58 AM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5 Re: How can I solve this limit problem What happens when y = x?
June 7th, 2011, 08:30 AM   #3
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Re: How can I solve this limit problem

Quote:
 Originally Posted by The Chaz What happens when y = x?
Then the limit will be =2 ? right?

thanks,

June 7th, 2011, 03:18 PM   #4
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Re: How can I solve this limit problem

Quote:
Originally Posted by abdlaziz
Quote:
 Originally Posted by The Chaz What happens when y = x?
Then the limit will be =2 ? right?

thanks,
On the other hand when y = -x, the limit =0. So it seems it is discontinuous at x=y=0.

 June 7th, 2011, 04:26 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Using the l'Hôpital's rule twice, $\lim_{(x,\,y)\to(0,\,0)}\,\frac{(x\,+\,y)^2}{x^2\, +\,y^2}\,=\,\lim_{(x,\,y)\to(0,\,0)}\,\frac{4\,+\, 4}{2\,+\,2}\,=\,\frac{8}{4}\,=\,2\text{.}$
 June 8th, 2011, 08:33 AM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: How can I solve this limit problem How did you apply L'Hopital's rule? What did you differentiate with respect to? As The Chas pointed out, when y= x, this becomes 4x^2/2x^2= 2 but when y-x it becomes 0/x^2= 0. That is, the limit is 2 approached along the line y= x but 0 approached along the line y= -x. The limit must be the same when approached along any line so this limit does not exist.
 June 8th, 2011, 09:22 AM #7 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5 Re: How can I solve this limit problem Welcome (back) to the forum, HallsofIvy! Also, it's "The Chaz"

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