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Simple multivariable limit problem

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]f(x,y) = 2x+3y[/itex]

    Let [itex]\epsilon[/itex] be any positive number. Show that there is a disk with center [itex](1,1)[/itex] and radius [itex]\delta[/itex] such that whenever [itex]P[/itex] is in that disk, [itex]\left| f(P) - 5\right| < \epsilon[/itex]. Give [itex]\delta[/itex] as a function of [itex]\epsilon[/itex].

    2. Relevant equations

    [itex]\left| 2x+3y - 5\right| < \epsilon[/itex]

    [itex]\sqrt{(x-1)^2 + (y-1)^2} < \delta[/itex]

    3. The attempt at a solution

    Obviously, it's a continuous equation that works out to exactly [itex]5[/itex] so the limit is 5. But I am stuck on solving for [itex]\delta[/itex] in terms of [itex]\epsilon[/itex]

    I suspect I am stuck on an easy step.
     
  2. jcsd
  3. Jan 24, 2013 #2

    HallsofIvy

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    [itex]|2x+ 3y- 5|= |2x- 2+ 3y- 3|= |2(x- 1)+ 3(y- 1)|\le 2|x- 1|+ 3|y- 1|[/itex]

    and, of course, [itex]|x- 1|\le \sqrt{(x- 1)^2+ (y- 1)^2}[/itex] and [itex]|y- 1|\le \sqrt{(x- 1)^2+ (y- 1)^2}[/itex].
     
  4. Jan 24, 2013 #3

    I like Serena

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    Welcome to PF, Toshe! :smile:

    Suppose we pick a point P with a distance r to (1,1).
    Then r < δ.

    What is the maximum value that |f(P)-5|=|2(x-1)+3(y-1)| can take as a function of r?
     
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