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Finding limit of function in 2 variables

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    lim as (x,y)-->(0,0) of sin(x^2+y^2)/(x^2+y^2)

    Questions: Does limit exist and if so, what is it.

    2. Relevant equations



    3. The attempt at a solution

    1. The professor instructed us to convert to polar coordinates to see if result depends on theta. If it does no limit.

    Conversion:

    lim as r-->0 of sin(r^2 cos^2(theta) + r^2 sin^2(theta)) / (r^2 cos^2(theta) + r^2 sin^2(theta))

    This I think reduces to sin(r^2)/r^2. This implies that the limit exists, correct(since the answer doesn't depend on theta)? However, this doesn't help me with actually finding the limit, as plugging in 0 still yields indeterminate 0/0. What do i do?
     
    Last edited by a moderator: Oct 8, 2007
  2. jcsd
  3. Oct 8, 2007 #2
    If you notice you can make a substitution!

    say u=x^2+y^2

    now your limit becomes:

    [tex]lim_{(x,y)->(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}[/tex]
    as (x,y)->(0,0) u->(0)
    [tex]lim_{u->0} \frac{\sin u}{u}[/tex]

    and well you should know sin u/u by heart xD

    also you could have done the substitution from sin(r^2)/r^2 actually it's almost the same thing since (x,y)->(0,0) (r,theta)->(0,0)

    and theta is nowhere in there so you have lim r->0 sin(r^2)/r^2
     
    Last edited: Oct 8, 2007
  4. Oct 8, 2007 #3
    i guess im just dumb, i didnt know that lim of sin u / u is 1. im taking math after a 2 year hiatus :(
     
  5. Oct 8, 2007 #4
    o well it's just a limit that sometimes comes up and when it does it's useful to know it. Also it's part of the proof of derivative of sin x.

    also when you have a limit like sin(r^2)/r^2 it's a calculus 1 type limit so you can use L'Hospital's rule if you get an indeterminate form, but using it here you get 0.
     
  6. Oct 8, 2007 #5
    dont u get 1? sin(r^2)/r^2 L'hopitaled becomes (cos(r^2)2r)/2r = cos(r^2) = 1?


    Anyway, thanks for help
     
    Last edited by a moderator: Oct 8, 2007
  7. Oct 8, 2007 #6
    o whops sorry i did the derivative 2x and got (2rsinr^2)/2r ><

    but yes you could do it multiple ways but the substitution would be fastest since you can do it straight away.
     
  8. Oct 9, 2007 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You don't need to use L'Hopital. You know from Calc I that
    [tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1[/tex]
    Just think of r2 as x.
     
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