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Delta/epsilon two variable limit

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data
    In calculus 1, honestly, I never understood this as it just doesn't make sense. Now in multivariable calculus, it still doesn't make sense. After all these engineering, math, and science classes, I still cannot figure these problems out, and it pisses me off!! I would appreciate it if someone could explain this example to me...

    Evaluate
    $$\lim_{(x,y)\rightarrow (0,0)} \frac{5x^2y}{x^2+y^2}$$


    2. Relevant equations



    3. The attempt at a solution
    This is an example problem, the book does not explain this well at all. The answer they get is: You can choose ##\delta = \varepsilon /5## and conclude that the limit = 0.

    I don't even see how they just (seemingly) magically conclude that.
    Any help is appreciated.
     
  2. jcsd
  3. Apr 7, 2014 #2

    LCKurtz

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    Try expressing the variables in polar coordinates. ##(x,y)\to (0,0)## is the same as ##r\to 0##.
     
  4. Apr 7, 2014 #3
    The definition of a limit would imply that the limit will be 0 if, for every small positive ε, there exists some δ such that |5xy2/(x2 + y2)| < ε whenever the distance from (x,y) to (0,0) is less than δ. We're can always ignore LARGE values of ε, so just assume ε < 1 (you'll see why).

    If x2+y2 < δ2, you know that x2 <= (x2+y2) < δ2 < δ; the last inequality depends on δ < ε < 1. So if δ = ε/5, then 5x2 < 5δ = ε. And you know that y/(x2+y2) is sin(θ), whose absolute value is <= 1. That proves the absolute value of the entire fraction is smaller than ε.
     
  5. Apr 7, 2014 #4

    LCKurtz

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    No, ##\frac y {\sqrt{x^2+y^2}}=\sin\theta##.
     
  6. Apr 7, 2014 #5
    So I can just change these functions into polar coordinates and solve that way?

    Is this the best way to approach these problems?
     
  7. Apr 7, 2014 #6

    LCKurtz

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    There is no best way. What you do depends on the nature of the problem. In this problem the degree of the numerator is 3 and the denominator is 2, plus the ##x^2+y^2## in the denominator suggest trying polar coordinates. Try it.
     
  8. Apr 7, 2014 #7

    Zondrina

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    This is also possible without polar co-ordinates. Start with the definition:

    $$∀ε>0,∃δ>0 \space | \space 0<||(x,y)-(0,0)||<δ⇒|\frac{5x^2 y}{x^2+y^2}-0|<ε$$

    Note that ##||(x,y)-(0,0)||<δ⇒|x|<δ∧ |y|<δ##.

    By the triangle inequality:

    ##|\frac{5x^2 y}{x^2+y^2}-0|= \frac{5|x|^2 |y|}{|x^2+y^2|} ≤ \frac{5|x|^2 |y|}{|x|^2+|y|^2 } < \frac{5δ^3}{2δ^2} < \frac{5δ^3}{δ^2} = 5δ##

    So we have:

    ##5δ ≤ ε ⇒ δ ≤ \frac{ε}{5}##

    Therefore, by choosing ##δ ≤ \frac{ε}{5}, |\frac{5|x|^2 |y|}{|x^2+y^2|}-0|<ε##. Although you should try the polar co-ordinates as well as it is often a useful trick. Try proving the chosen ##\delta## does indeed satisfy the definition.
     
    Last edited: Apr 7, 2014
  9. Apr 7, 2014 #8

    jbunniii

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    Perhaps simpler:

    If ##x = 0## and ##y \neq 0##, then we have
    $$\frac{5x^2y}{x^2+y^2} = 0$$

    If ##x \neq 0## we may write
    $$\frac{5x^2y}{x^2+y^2} = \frac{5y}{1 + y^2/x^2}$$
    As ##y \rightarrow 0##, the numerator goes to zero. The denominator is at least ##1## for all ##x,y##. What can you conclude?
     
  10. Apr 8, 2014 #9

    LCKurtz

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    If we are going to explore alternatives, you can also use ##|xy|\le \frac{x^2+y^2} 2## which comes from ##(|x|-|y|)^2\ge 0##. Then$$
    \left|\frac{5x^2y}{x^2+y^2}\right|\le 5 |x|\left|\frac{xy}{x^2+y^2}\right|\le\frac 5 2 |x|$$which leads to a solution too. So @iraid, I guess you have seen more than one way to skin this particular cat. Still, I think polar coordinates is the simplest.
     
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