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Proving that multivariable limit exists

  1. Aug 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Given f(x,y)=(y+x)/(y-x) use an ε-∂ proof to show that lim(x,y)→(0,1) f(x,y) exists.

    2. Relevant equations
    |(y+x)/(y-x)-1|=|(2x)/(y-x)|

    3. The attempt at a solution
    I know that the limit is 1. I cant figure out how to massage the above any further to get it into the form |(2x)/(y-x)|<=k|x| or |(2x)/(y-x)|<=k|y-1| so that I can choose an appropriate value for ∂. I have tried restricting ∂<1 but it doesn't get me any further. Any hints would be appreciated.

    Thanks.
     
  2. jcsd
  3. Aug 18, 2015 #2

    Ray Vickson

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    In problems of this type is often useful to translate the desired point to the origin. By writing ##x = x'## and ##y = 1 + y'##, the limit becomes ##(x',y') \to (0,0)##. Try it and see what happens.
     
  4. Aug 18, 2015 #3

    andrewkirk

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    Saying the limit exists and equals 1 is saying that ##\forall\epsilon>0\ \exists\delta>0## such that ##d\big((x,y),(0,1)\big)<\delta\Rightarrow |f(x,y)-1|<\epsilon##, where ##d\big(x1,y1),x2,y2)\big)\equiv\sqrt{(x1-x2)^2+(y1-y2)^2}## is the distance function.

    So what you need to do is show how to find a ##\delta## that works, given ##\epsilon##.

    The problem is easier to do if you concentrate on a square in the number plane centred on (0,1), rather than the circle defined by the distance formula. So, given ##\epsilon##, try to find ##\alpha>0##, defined in terms of epsilon, such that if ##|x-0|<\alpha## and ##|y-1|<\alpha## then ##|f(x,y)-1|<\epsilon##. That will be a matter of simple arithmetic using the formulas you have written above.

    If you can do that, you only need to find ##\delta## such that the circle around (0,1) defined by ##\delta## contains the square around (0,1) defined by ##\alpha## so that
    ##d\big((x,y),(0,1)\big)<\delta\Rightarrow |x-0|<\alpha\wedge|y-1|<\alpha## .
     
  5. Aug 19, 2015 #4

    Zondrina

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    You got to the point:

    $$\frac{2 |x|}{|y - x|} < \varepsilon$$

    Using the reverse triangle inequality, you get:

    $$\frac{2 |x|}{|y - x|} \leq \frac{2 |x|}{\left| |y| - |x| \right|} \leq \frac{2 |x|}{|y| - |x|} < \varepsilon$$

    You know ##|x| < \delta## and ##|y - 1| < \delta##. Use these inequalities to put a bound on ##\frac{1}{|y| - |x|}## in terms of ##\delta##. This will let you solve for ##\delta( \varepsilon)##.
     
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