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Having trouble applying squeeze theorem

  1. Jul 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the limit, if it exists, or show that the limit does
    not exist:

    lim (x,y) -> (0,0) [(y^2*sin(x)^2)/(x^4+y^4)]

    (According to the textbook the limit 'does not exist')


    3. The attempt at a solution

    Since the function is approaching the origin [(0,0)]:

    test path along y-axis:
    let x = 0
    lim (y) -> (0) [0/y^4] = 0

    test path along x-axis:
    let y = 0
    lim (x) -> (0) [0/x^4] = 0

    This is not efficient enough to prove that the limit is 0, therefore investigate further possibly using the squeeze theorem.

    I know that [(y^2)/(x^4+y^4)] <= 1 (but i don't know what to do with this information)

    0 <= [(y^2*sin(x)^2)/(x^4+y^4)] <= **

    ** Here is where i got stuck. I do not know what function fits the criteria and how to look for it. Please excuse me I just learned this theorem recently and I'm hoping someone can help me.
     
  2. jcsd
  3. Jul 29, 2014 #2
    So you now know that the limit is zero if you approach ##(0,0)## from the x-axis (or y-axis). Can you find another path that will give you a nonzero limit?
     
  4. Jul 29, 2014 #3
    So far I've tried:

    let y = x
    let y = x^2
    let y = x^3
    let y = sqrt(x)

    and got 0 for all. Is it possible to solve using the squeeze theorem?
     
  5. Jul 29, 2014 #4
    Try ##y=x## again and show your work.
     
  6. Jul 29, 2014 #5
    let y = x

    lim (x,x) -> (0,0) [(x^2*sin(x)^2)/(x^4+x^4)]
    = lim (x) -> (0) [(x^2*sin(x)^2)/(2*x^4)]
    = lim (x) -> (0) [(x^2*sin(x)^2)/(2*x^2*x^2)]
    = lim (x) -> (0) [(sin(x)^2)/(2*x^2)]
    = 0
     
  7. Jul 29, 2014 #6
    Have another look at your last equality. Are you sure that
    \begin{equation*}
    \lim_{x \rightarrow 0} \frac{ \sin^2 x}{2x^2} = 0?
    \end{equation*}
     
  8. Jul 29, 2014 #7
    Yes you're right I would get 0/0 (undefined) and the limit does not exist. The other paths I mentioned early turn out to be the same. Thanks for your help. Could this be solved using the squeeze theorem?
     
  9. Jul 30, 2014 #8
    No. The limit
    \begin{equation*}
    \lim_{x \rightarrow 0} \frac{\sin^2 x}{2x^2}
    \end{equation*}
    DOES exist. It is just not zero! You should compute it. You could use the fact that you know (?) ##\lim_{x \rightarrow 0} \frac{\sin x}{x}## or use something powerful like a Taylor expansion or l'Hopital.
     
  10. Jul 30, 2014 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
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    Not really. The squeeze theorem allows you to prove that a function has a limit, but in this case, you're trying to show that the limit of a given function does not exist.
     
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