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: Two variable limits

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data

    The real valued function f of two variables is defined by f(x,y) =
    tan( (1/2) *(pi) *sqrt( (x^2 + y^2) )/( sqrt( (x^2 + y^2) ) for each (x,y) satisfying 0 < x^2 + y^2 < 1.

    How should f(0,0) defined so that f is continuous at (0,0)?

    2. Relevant equations

    3. The attempt at a solution
    I wasn't sure whether the question was asking show the limit exists or doesn't exist? I think asking me to show it exists.
    I tried using polar co ordinates
    y=r*sin[tex]\theta[/tex] , x=r*cos[tex]\theta[/tex]
    =tan( (1/2) * (pi) * r) / r
    = 1/cos( (1/2) * (pi) * r ) * sin((1/2) * (pi) * r)/r
    = 1 * sin((1/2) * (pi) * r)/r

    I got stuck here was trying to use the identity lim->0 sinx/x = 1
    I know you can use squeeze theorem to prove limits exist, but I don't really understand how it works for two variables.
    In one variable limits if 0 < |x-a|<[tex]\delta[/tex] then |f(x)-L|< [tex]\epsilon[/tex]
    but in two variable limits what is |x-a|?
  2. jcsd
  3. Apr 1, 2009 #2


    User Avatar
    Science Advisor

    since sin(x)/x-> 1, for [itex]sin(\pi r/2)/r, let [/itex]x= (\pi/2)r[itex]. What is r equal to in terms of x? Replace the r in the denominator by that.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook