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 $\epsilon>0$ there exist $\delta> 0$ 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!