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Easy delta/epsilon proof of a multivariable limit

  1. Jan 18, 2012 #1

    I'm trying to wrap my head around epsilon/delta proofs for multivariable limits and it turns out I became stuck on an easy one!

    The limit is:

    [itex]\lim_{(x,y) \to (1,1)}\frac{xy}{x+y}[/itex]

    Obviously, the result is [itex]1/2[/itex], but I'm unable to prove it!

    Any hints?

    Thank you!
  2. jcsd
  3. Jan 19, 2012 #2


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    Well, what have you done? I assume you want to prove that

    "Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x-1)^2+ (y-1)^2}<\delta[/itex] then
    [tex]\sqrt{\frac{xy}{x+y}- 1}<\epsilon[/tex]"
  4. Jan 19, 2012 #3
    Actually I think I might have solved it:

    I want to prove that

    Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x-1)^2+ (y-1)^2}<\delta[/itex] then
    [tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<\epsilon[/tex]


    [tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|=\left|\frac{2xy-(x+y)}{2(x+y)}\right|=\left|\frac{xy + xy-x-y)}{2(x+y)}\right|=\left|\frac{x(y-1)+y(x-1)}{2(x+y)}\right|\leq\left|\frac{x(y-1)}{2(x+y)}\right|+\left|\frac{y(x-1)}{2(x+y)}\right|[/tex] by the triangle inequality.

    Now, I know that since [itex]{(x,y) \to (1,1)}[/itex] I can make [itex]x[/itex] and [itex]y[/itex] positive so that [itex]\left|\frac{x}{x+y}\right|\leq1[/itex] and [itex]\left|\frac{y}{x+y}\right|\leq1[/itex]. [I'm no sure about this argument]

    This way,


    But this is just a square with side 2 around the point [itex](x, y)[/itex]. This square is entirely contained by the circle with center [itex](x, y)[/itex] and radius [itex]\sqrt{2}[/itex].

    So, when given some [itex]ε>0[/itex] if I set [itex]δ=min(1, ε\sqrt{2})[/itex] and I choose a point [itex](x, y)[/itex] such that [itex]\sqrt{(x-1)^2+ (y-1)^2}<ε\sqrt{2}[/itex], then certainly [itex]|y-1|+|x-1|<ε[/itex] and, by extension,

    [tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε[/tex]

    What's wrong with my rationale? :-)
    Last edited: Jan 19, 2012
  5. Jan 19, 2012 #4


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    I did not read anything other than this line so I cannot comment on anything else you wrote, but this is wrong. Take (x,y) such that ε < [(x-1)2+(y-1)2]1/2 ≤ ε21/2. Then notice that [(x-1)2+(y-1)2]1/2 ≤ |x-1|+|y-1| to find (x,y) with [(x-1)2+(y-1)2]1/2 ≤ ε21/2 but ε < |x-1|+|y-1|.
  6. Jan 19, 2012 #5
    Obviously, you are right (thank you). Take [itex]ε=1[/itex]. I can choose a point in the circle with radius [itex]\sqrt{2}[/itex] that is not inside the square [itex]|x-1|+|y-1|<1[/itex].

    How about this: if I choose a point inside the circle with radius ε, then that point is also inside the square with side 2ε, because the circle is contained within the square.

    Therefore, if I set [itex]δ\leq ε[/itex], then I know that any point inside the circle with radius δ verifies [itex]|x-1|+|y-1|<ε[/itex], which, if the first part of my derivation is correct, implies that

    [tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε[/tex]

    Is this correct?
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