# Prove Lim (x,y)->(1,2) (x + y) = 3.

• Sep 8th 2010, 08:19 AM
tarheelborn
Prove Lim (x,y)->(1,2) (x + y) = 3.
Consider R^2 where p((a,b),(c,d))=Sqrt((a-c)^2+(b-d)^2) and R where p(x,y) = |x-y|. f:R^2->R f((x,y))=x+y. Prove Lim (x,y)->(1,2) (x + y) = 3.

I really don't know quite where to start this problem. I know the limit will exist, if it exists, in R. I think I might need to know the distance from (x,y) to (1,2). But then where would I go with that? Thanks for your help.
• Sep 8th 2010, 10:00 AM
TheEmptySet
Note that
$p[(x,y),(1,2)]=\sqrt{(x-1)^2+(y-2)^2}< \delta$
This implies that

$(x-1)^2+(y-2)^2 < \delta^2$ and that

$(x-1)^2< \delta^2 \implies |x-1|< \delta$
$(y-2)^2< \delta^2 \implies |y-2|< \delta$

Now just let $\epsilon > 0$ and choose $\delta =$...
• Sep 8th 2010, 02:31 PM
tarheelborn
Would you not need to choose two deltas? I just don't see where to take this.
• Sep 8th 2010, 02:56 PM
Defunkt
$|f(x,y) - 3| = |x + y -3| = |x - 1 + y - 2| \le |x-1| + |y-2|$
• Sep 8th 2010, 03:36 PM
HallsofIvy
Here's a rather different way to do it: Let u= x- 1, v= y- 2. Then x= u+ 1, y= v+ 2 so x+ y= u+ v+ 3 and the problem becomes $\lim_{(u,v)\to (0, 0)}u+ v+ 3= 3$.

Now convert to polar coordinates: $u= r cos(\theta)$, $v= r sin(\theta)$ so that $u+ v+ 3= r(cos(\theta)+ sin(\theta))+ 3$. The point of that is that now the single variable "r" measures the distance from (u, v) to (0, 0). For any $\theta$, $|cos(\theta)|$ and $|sin(\theta)|$ are both less than or equal to 1 so $|cos(\theta)+ sin(\theta)|$ is less than or equal to 2. $|r(cos(\theta)+ sin(\theta)+ 3 -3|\le 2|r|$.
• Sep 8th 2010, 05:35 PM
tarheelborn
Quote:

Originally Posted by Defunkt
$|f(x,y) - 3| = |x + y -3| = |x - 1 + y - 2| \le |x-1| + |y-2|$

But did we lose the 3 in the last part of this equation? Or do I just have |x-1|<\delta/2 and so on?
• Sep 8th 2010, 05:39 PM
Defunkt
Quote:

Originally Posted by tarheelborn
But did we lose the 3 in the last part of this equation? Or do I just have |x-1|<\delta/2 and so on?

No, and you only need one delta: as TES showed,
$\rho \left( (x,y), (1,2)) < \delta \Rightarrow \ldots \Rightarrow |x-1| < \delta$ and $|y-2| < \delta$

so $|f(x,y) - 3| \le |x-1| + |y-2| < 2 \delta$
• Sep 8th 2010, 06:13 PM
tarheelborn
So would you say that delta = epsilon/2 so when you add them you get epsilon?
• Sep 8th 2010, 06:21 PM
Defunkt
Yep.