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November 17th, 2009, 03:46 AM   #1
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prove the limit exist (Partial Differential)

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
x,y,z-->(0,0,0)

can anyone guide me solving this question..
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November 17th, 2009, 02:40 PM   #2
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Re: prove the limit exist (Partial Differential)

Remember L'Hopital's rule? It's basically the same story here, just in multiple dimensions.
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November 17th, 2009, 04:26 PM   #3
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Re: prove the limit exist (Partial Differential)

ok.. 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?

ok.. i proceed..
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 to prove the limit is exist or not?
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November 17th, 2009, 09:44 PM   #4
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Re: prove the limit exist (Partial Differential)

The limit of f/g = the limit of f'/g' if it exists.

Your transformation to a single variable is ok in this case because of the symmetry of the functions f and g, but bear in mind that what you have to show in general is that the limit of the ratio of partials exists and is the same for each of the partial derivatives d/dx d/dy d/dz. This way, the limit will be the same no matter what path you choose to approach the origin.
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November 22nd, 2009, 06:21 PM   #5
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Re: prove the limit exist (Partial Differential)

ok rival...
so.. is my calculations above is correct?
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November 22nd, 2009, 09:39 PM   #6
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Re: prove the limit exist (Partial Differential)

yes
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