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Prove the limit exist (Partial Differential)

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi everyone this is my first time here - I'm completely confused not so I thought I may as well ask complete stangers for help!

    the question is..
    Determine whether the following limit exists. If so, find its value.

    lim [sin(x^2+y^2+z^2)] / (x^2+y^2+z^2)^1/2

    2. Relevant equations
    sin x/x = 1

    3. The attempt at a solution
    ok this is my idea..
    if i let t=x^2+y^2+z^2, the limit should be like..

    lim (sin t) / (t^1/2)

    apply L'Hospital rule,

    (cos t) / (t^-1/2) = t^1/2 cos t

    am i right?

    t=0.. so...

    (cos t) / (t^-1/2) = t^1/2 cos t
    = 0^1/2 cos 0
    = 0
    ok.. how am i going to proceed?
  2. jcsd
  3. Nov 22, 2009 #2


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    Homework Helper

    first be a little careful, if you let t=x^2+y^2+z^2, then you are only looking at the limit form t->0+

    that said i think your approach is valid, though you have missed a factor of -2 in your L'Hop differentiation, but you still get the limit is zero...

    to see what is going on, if you let r = (x^2+y^2+z^2)^(1/2), you are essentially changing to spherical co-ordinates, where the limit is independent of theta or phi
    [tex] \lim{r \to 0^+} \frac{sin(r^2)}{r} [/tex]

    as another method, how about multiplying through by r/r then considering the limit ?
  4. Nov 22, 2009 #3
    hurmm.. ok..
    so.. wut bout limit for 0-?
  5. Nov 22, 2009 #4


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    Homework Helper

    i was just trying to clarify for your variable change you only consider t->0+, so you don't have to worry about 0-. Its not hugely important here, just something to be aware of
  6. Nov 22, 2009 #5
    ok... let me refresh back my calculation...

    lim ...............[sin(x^2+y^2+z^2)] / (x^2+y^2+z^2)^1/2

    apply L'hospital rule..
    so.. i need to differentiate nominator, f(x) and denominator, g(x)
    let r = (x^2+y^2+z^2)^(1/2)
    so.. i shall have sin r^2/r

    then, (2r cos r^2)/1

    substitute limit r=0
    i'll get 2(0) cos 0
    then the answer is 0..
    so.. the limit is exist and the value is 0..

    can u please help to confirm my answer lanedance..
  7. Nov 25, 2009 #6
    can we apply a conjugate into this question?

    the answer will be:
    you'll get the sin x/x = 1 thingy.

    then your answer will be left with:
    lim.................(x^2 + y^2 + z^2)^(1/2)

    and yada, yada, yada...
  8. Nov 25, 2009 #7
    damn bro/sis
    r u one of mmu's cyberjaya student?
    u got the answer for other question?
    damn what a coincidence
  9. Nov 25, 2009 #8
    how about the other questions?
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