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Delta-Epsilon Proof of a Limit with 2 Variables

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove using the formal definition of a limit that
    [tex]\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2[/tex]
    is equal to 1.
    2. Relevant equations
    [tex]\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2\\
    \left \| \overline{x}-\overline{a} \right \|< \delta
    \\
    \left | f(\overline{x})-L \right |<\epsilon[/tex]

    3. The attempt at a solution
    [tex]\sqrt{(x-1)^2+(y-2)^2}<\delta\\
    \left | 5x^3-x^2y^2-1 \right |<\epsilon[/tex]
    I have no idea where to go from there. I can't figure out how to manipulate the second equation to resemble the first.
     
  2. jcsd
  3. Mar 8, 2013 #2

    Zondrina

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    Perhaps I don't have the best ( biggest or smallest depending on what you want ) possible neighborhood with the delta I found, but :

    ##δ = min \{ 1, \sqrt[4]{ε/6} \}## will suffice I believe.

    Use the fact that ##0 < |x-1|, |y-2| < δ## to deduce it and possibly get an even better delta ( Trust me I was being lazy here, you can probably do much much better than the value I've given ).
     
    Last edited: Mar 8, 2013
  4. Mar 8, 2013 #3

    LCKurtz

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    I wouldn't try to make it resemble the first. Try thinking about it this way. You do know that you can make both ##|x-1|## and ##|y-2|## small. And you are trying to make$$
    |5x^3-x^2y^2-1|$$small. Now, you know that ##5x^3## is going to get close to ##5##, so let's subtract and add ##5##:$$
    |5x^3-5 + 5-x^2y^2-1| = |5(x^3-1)+4-x^2y^2|$$Now that ##x^2y^2## term is going to get close to ##4x^2## as ##y## gets close to ##2##, so lets add and subtract ##4x^2##:$$
    |5(x^3-1)+4-x^2y^2 - 4x^2+4x^2|=|5(x^3-1)+4(1-x^2)+x^2(4-y^2)|$$Now do you see how to make it small?
     
    Last edited: Mar 8, 2013
  5. Mar 8, 2013 #4
    so you end up with this?
    [tex]|5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|<\epsilon[/tex]
    I see that it contains 3 terms with factors less than delta but I must be missing a some concept here.
     
  6. Mar 8, 2013 #5

    Zondrina

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    If you want to follow the path that LC is taking, apply the triangle inequality to what you have right now.
     
  7. Mar 8, 2013 #6

    LCKurtz

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    Yes, and each term has something you can make small multiplied by factors you can keep bounded.
     
  8. Mar 11, 2013 #7
    Does that mean
    [tex]5(x^2+x+1)(x-1)-4(x-1)|x+1|-x^2(y-2)|y+2|<\epsilon[/tex]
    I don't understand.
     
  9. Mar 11, 2013 #8

    LCKurtz

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    Actually, you don't have it less than ##\epsilon## yet; that is your goal. What you do have is that$$
    |5x^3-x^2y^2-1|\le |5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|$$
    That isn't what the triangle inequality gives you. It says ##|a + b| \le |a| + |b|## so you would get$$
    |5x^3-x^2y^2-1|\le |5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|$$ $$\le 5|x-1| |x^2+x+1|+4|x-1| |x+1|+|x^2||y-2||y+2|$$You want to make the whole thing less than ##\epsilon##, which you could do by making each term less than ##\epsilon/3##. You know you can make both ##|x-1|## and ##|y-2|## as small as you want. So if the other terms don't get too big you should be able to make it work. Remember ##x## and ##y## can't be just anything because they are getting close to ##1## and ##2##, respectively. So say you make sure ##\delta < 1## so both x and y are within 1 of their limits. Then how big can the extra x and y expressions be? Overestimate some more then figure out what ##\delta## will work.
     
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