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

**tarheelborn** · September 8th 2010, 07:19 AM

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.

**TheEmptySet** · September 8th 2010, 09:00 AM

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 $\delta =$ ...

**tarheelborn** · September 8th 2010, 01:31 PM

Would you not need to choose two deltas? I just don't see where to take this.

**Defunkt** · September 8th 2010, 01:56 PM

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

**HallsofIvy** · September 8th 2010, 02:36 PM

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. $|r(cos(\theta)+ sin(\theta)+ 3 -3|\le 2|r|$ .

**tarheelborn** · September 8th 2010, 04:35 PM

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?

**Defunkt** · September 8th 2010, 04:39 PM

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 $|f(x,y) - 3| \le |x-1| + |y-2| < 2 \delta$

**tarheelborn** · September 8th 2010, 05:13 PM

So would you say that delta = epsilon/2 so when you add them you get epsilon?

**Defunkt** · September 8th 2010, 05:21 PM

Yep.

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