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[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

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- #1

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[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

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Would you show us what you already tried?

- #3

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Start off with the definition:Stevecgz said:

[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

[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]

[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]

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arildno

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Prove that the sum of two continuous functions is itself a continuous function.

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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

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HallsofIvy

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