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

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    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.
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  2. #2
    Behold, the power of SARDINES!
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    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 = ...
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  3. #3
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    Would you not need to choose two deltas? I just don't see where to take this.
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    |f(x,y) - 3| = |x + y -3| = |x - 1 + y - 2| \le |x-1| + |y-2|
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  5. #5
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    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|.
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    Quote Originally Posted by Defunkt View Post
    |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?
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    Quote Originally Posted by tarheelborn View Post
    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
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    So would you say that delta = epsilon/2 so when you add them you get epsilon?
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  9. #9
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    Yep.
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