Results 1 to 6 of 6

Math Help - Continuity of a 2 variables function

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    13

    Continuity of a 2 variables function

    Hello,

    I need a hand with an exercise, it asks me to study the continuity of the following function for each value of "a" (sorry I don't know how to use LaTex!):


    f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if y != 0
    f(x,y) = a, if y = 0

    Basically I don't how to solve the limit for y -> 0

    Any idea?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,357
    Thanks
    36
    Quote Originally Posted by jollysa87 View Post
    Hello,

    I need a hand with an exercise, it asks me to study the continuity of the following function for each value of "a" (sorry I don't know how to use LaTex!):


    f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if y != 0
    f(x,y) = a, if y = 0

    Basically I don't how to solve the limit for y -> 0

    Any idea?

    Thanks in advance.
    If what you're asking is

     <br />
\lim_{(x,y) \to (0,0)} \frac{x^2 e^x}{y \,e^{y^2}}<br />

    try letting x = r \cos \theta and y = r \sin \theta and then let r \to 0 .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2009
    Posts
    13
    I already did it but the problem is that x can be anything, otherwise the exercise would have been:

    f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if (x,y) != 0
    f(x,y) = a, if (x,y) = 0

    Am I wrong??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,547
    Thanks
    1395
    So the problem is not \lim_{(x,y)\rightarrow (0,0)} \frac{x^2e^x}{ye^{y^2}}, it is \lim_{y\rightarrow 0}\frac{x^2e^x}{ye^{y^2}}?

    Well, that's easier! If x is not 0, the denominator goes to 0 while the numerator does not so there is no limit. If x= 0, the function is 0 for all y and so the limit is 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2009
    Posts
    13
    so i can say that the limit for x different from 0 doesn't exist simply because the denominator goes to 0 while the numerator does not or I have to make some better explanation? thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,357
    Thanks
    36
    Quote Originally Posted by jollysa87 View Post
    so i can say that the limit for x different from 0 doesn't exist simply because the denominator goes to 0 while the numerator does not or I have to make some better explanation? thanks!
    Yes, that is sufficient. The case where both x and y approach zero is really the only interesting case.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 18th 2011, 02:27 AM
  2. Continuity of a function of 2 variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 02:11 PM
  3. Continuity of a 2 variables function in R2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 16th 2009, 05:17 AM
  4. Continuity in 2 variables.
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 22nd 2009, 11:12 PM
  5. Function in two variables - continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 21st 2009, 10:45 PM

Search Tags


/mathhelpforum @mathhelpforum