Results 1 to 4 of 4

Math Help - Limit and Continuity (multivariable calculus) [help]

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    7

    Exclamation Limit and Continuity (multivariable calculus) [help]

    Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.
    i.e. for every point (a,b) in the plane and for every epsilon>0, find delta>0 depending on a,b and epsilon such that whenever
    0<|(x,y)-(a,b)|<delta, |f(x,y)-f(a,b)|<epsilon

    How to find an epsilon in terms of delta? I am troubled in expressing epsilon as a function of delta.

    Many thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by catcat103 View Post
    Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.
    i.e. for every point (a,b) in the plane and for every epsilon>0, find delta>0 depending on a,b and epsilon such that whenever
    0<|(x,y)-(a,b)|<delta, |f(x,y)-f(a,b)|<epsilon

    How to find an epsilon in terms of delta? I am troubled in expressing epsilon as a function of delta.

    Many thanks.
    epsilon is an arbitrary number.. you should find delta in terms of epsilon and not the other way around..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by catcat103 View Post
    Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.
    We shall prove f is continous at (a,b). Let \sqrt{(x-a)^2+(y-b)^2} < \delta. Then |x-a| = \sqrt{(x-a)^2}\leq \sqrt{(x-a)^2+(y-b)^2} < \delta. Similarly |y-b| < \delta. Now use the inequality ||y|-|b|| \leq |y-b| < \delta \implies - \delta < |y| - |b| < \delta. Thus, |y| < \delta + |b|. With these facts we can prove continuity. We need to show |xy-ab| can be made arbitrary small. Add and subtact, |y(x-a) + a(y-b)| \leq |y||x-a| + |a||y-b| < (\delta + |b|)\delta + |a| \delta < \delta (1+|a|+|b|) if 0<\delta \leq 1.
    This tells us that for any \epsilon > 0 choose \delta  = \min \left\{ 1 , \tfrac{\epsilon}{1+|a|+|b|} \right\}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2008
    Posts
    7
    ah...i forgot those useful inequalities. what a marvellous answer! thanks for your reply!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Continuity of multivariable function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2009, 12:25 PM
  2. continuity of multivariable functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 10th 2009, 01:37 AM
  3. Limit for multivariable calculus
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 16th 2008, 06:22 PM
  4. Replies: 1
    Last Post: June 23rd 2008, 09:17 AM
  5. Replies: 1
    Last Post: February 26th 2006, 02:50 PM

Search Tags


/mathhelpforum @mathhelpforum