1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    \lim_{x \rightarrow 0} \frac{ \sin^2 x}{2x^2} = 0?
  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
    \lim_{x \rightarrow 0} \frac{\sin^2 x}{2x^2}
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted