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 February 13th, 2014, 10:00 AM #1 Member   Joined: Feb 2014 Posts: 30 Thanks: 0 Multivariable Calculus problem $\lim _{ (x,y)\rightarrow (0,0) }{ ((x-y)cos(\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } ) }$ I set -1<=cos<=1 and get the limit of the whole thing is 0. But it look like it may be DNE on the graph. Can someone help me? $\lim _{ (x,y)\rightarrow (0,0) }{ (\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }+{ y }^{ 2 } } ) }$ This looks like it has limit =0 but I'm not sure, I tried approaching 0,0 in different ways and they came out to be 0. But I don't know how to prove that it is. Thank you
 February 13th, 2014, 10:37 AM #2 Member   Joined: Feb 2014 Posts: 30 Thanks: 0 Re: Multivariable Calculus problem Please help me
 February 13th, 2014, 12:16 PM #3 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Re: Multivariable Calculus problem Have you tried to express the functions in polar coordinates? In that way it should be easier to see what's going on.
 February 13th, 2014, 01:28 PM #4 Member   Joined: Feb 2014 Posts: 30 Thanks: 0 Re: Multivariable Calculus problem $({ r }^{ 2 }{ cos }^{ 2 }\theta \quad -\quad { r }^{ 2 }{ sin }^{ 2 }\theta )/({ r }^{ 2 }{ sin }^{ 2 }\theta \quad +\quad rcos\theta )$ I don't think it's gonna go anywhere
February 13th, 2014, 01:32 PM   #5
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Re: Multivariable Calculus problem

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The limit is 0. (What is DNE?)

February 13th, 2014, 02:10 PM   #6
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Re: Multivariable Calculus problem

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Originally Posted by mathman
Quote:
The limit is 0. (What is DNE?)
It is does not exist.
Thank you

 February 13th, 2014, 04:08 PM #7 Member   Joined: Feb 2014 Posts: 30 Thanks: 0 Re: Multivariable Calculus problem Can you show me how to do it please?
February 15th, 2014, 01:15 PM   #8
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Re: Multivariable Calculus problem

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 Originally Posted by maluita659 Can you show me how to do it please?
Your original proof is correct. You have a product of two terms where one -> 0 and the other has absolute value bounded by 1. The product -> 0.

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