Epsilon-delta proof

  • Thread starter Stevecgz
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I'm working on an Epsilon-delta proof for:

[tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3[/tex]

I've tired a couple of different approaches but am not getting anywhere (I'm terrible at these). Any suggestions would be appreciated. Thanks.

Steve
 

Answers and Replies

  • #2
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Would you show us what you already tried?
 
  • #3
benorin
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Stevecgz said:
I'm working on an Epsilon-delta proof for:

[tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3[/tex]

I've tired a couple of different approaches but am not getting anywhere (I'm terrible at these). Any suggestions would be appreciated. Thanks.

Steve
Start off with the definition:

[tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3 \mbox{ if, and only if }[/tex]
[tex]\mbox{for every }\epsilon >0,\mbox{ there exists a }\delta >0 \mbox{ such that } 0<\sqrt{(x-1)^2+(y-1)^2}<\delta \mbox{ implies that }|x^2 + xy + y - 3| <\epsilon[/tex]​
 
  • #4
arildno
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Prove that the sum of two continuous functions is itself a continuous function.
 
  • #5
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benorin said:
Start off with the definition:

[tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3 \mbox{ if, and only if }[/tex]
[tex]\mbox{for every }\epsilon >0,\mbox{ there exists a }\delta >0 \mbox{ such that } 0<\sqrt{(x-1)^2+(y-1)^2}<\delta \mbox{ implies that }|x^2 + xy + y - 3| <\epsilon[/tex]​

Thanks benorin. I was trying with [tex]0<\sqrt{x^2+y^2}<\delta[/tex]. I think I can get it now.

Steve
 
  • #6
HallsofIvy
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By the way, while [itex]\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2}[/itex] is the "standard metric" it would be equivalent to show the "for every [itex]\epsilon>0[/itex] there exist [itex]\delta> 0[/itex] such that max (|x-1|,|y-1|)< [itex]\delta[/itex] implies that |x2+ xy+ y- 3|<[itex]\epsilon[/itex] and might be simpler. Of course, your teacher might require that you be able to show that they are equivalent!
 

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