# Epsilon-delta proof

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

Would you show us what you already tried?

benorin
Homework Helper
Stevecgz said:
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
Homework Helper
Gold Member
Dearly Missed
Prove that the sum of two continuous functions is itself a continuous function.

benorin said:
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!