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Homework Help: Limit problem

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Apply the definition of the limit to show that

    lim (x,y)-->(0,0) xy^3/(x^2+y^2) = 0


    2. Relevant equations
    Definition of the limit:
    lim (x,y)-->(a,b) f(x,y) = L if for every number epsilon > 0 there is a corresponding number delta > 0 such that if (x,y) is in the domain and 0 < sqrt((x-a)^2 + (y-b)^2) < delta then |f(x,y) - L| < epsilon

    3. The attempt at a solution

    So far I've just plugged in the numbers:

    Let epsilon > 0. We want to find delta > 0 such that
    if 0 < sqrt(x^2 + y^2) < delta then |xy^3/(x^2 + y^2) - 0| < epsilon

    I have no idea what to do next.
    Note that it says to show that this is the limit using the definition given above.
     
  2. jcsd
  3. Aug 15, 2011 #2
    Is not true that the limit is 0.
    If you let [itex]y=0[/itex] and [itex]x \to 0[/itex], we have the
    [tex]\lim_{x \to 0} \frac{1}{x} \ne 0[/tex]
     
  4. Aug 15, 2011 #3
    If y = 0 then xy^3 = 0 then xy^3/(x^2+y^2) = 0/x^2 = 0 so the limit must be 0
     
  5. Aug 15, 2011 #4
    u cant take limit of the numerator without taking limit of the denominator so u basically wrong
     
  6. Aug 15, 2011 #5
    Wolfram?
    http://www.wolframalpha.com/input/?i=lim+%28x%2Cy%29-%3E%280%2C0%29+xy^3%2F%28x^2+%2B+y^2%29
     
  7. Aug 15, 2011 #6
    sry my bad :D
     
  8. Aug 15, 2011 #7
    Oh yes, sure.
     
  9. Aug 15, 2011 #8

    HallsofIvy

    User Avatar
    Science Advisor

    I would switch to polar coordinates- that way the "distance to (0, 0)" is given by the single variable, r.

    In polar coordinates, the function is
    [tex]\frac{(r cos(\theta)(r sin(\theta))^3}{r^2}= r^2 cos(\theta)sin^3(\theta)[/tex]

    Of course, [itex]|cos(\theta)sin^3(\theta)|\le 1[/itex] for all [itex]\theta[/itex].
     
  10. Aug 16, 2011 #9
    Ok let me know if this makes any sense at all....

    After changing it to polar coordinates, we get [itex] \lim_{r \to 0} r^2 cos(\theta)sin^3(\theta)[/itex]

    Let [itex]\epsilon > 0[/itex]. We want to find [itex]\delta > 0[/itex] such that

    if [itex] 0<r< \delta[/itex] then [itex]|r^2 cos(\theta)sin^3(\theta) - 0| < \epsilon [/itex]

    since [itex]|cos(\theta)sin^3(\theta)| \le 1[/itex] for all [itex]\theta[/itex]

    [itex]r^2 cos(\theta)sin^3(\theta) \le r^2[/itex]

    Thus if we choose [itex]\delta^2 = \epsilon[/itex] and let [itex]0<r<\delta[/itex] then

    [itex]|r^2 cos(\theta)sin^3(\theta) - 0| \le r^2 \le \delta^2 = \epsilon[/itex]

    And hence [itex] \lim_{(x,y) \to (0,0)} = 0 [/itex]
     
    Last edited: Aug 16, 2011
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