Epsilon-Delta Proof: Limiting x^2 + xy + y to 3

[Stevecgz]

I'm working on an Epsilon-delta proof for:

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

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

 

[devious_]

Would you show us what you already tried?

 

[benorin]

I'm working on an Epsilon-delta proof for:

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

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:

\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3 \mbox{ if, and only if }
\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​

 

[arildno]

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

 

[Stevecgz]

Start off with the definition:

\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3 \mbox{ if, and only if }
\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​

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

Steve

 

[HallsofIvy]

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