Results 1 to 7 of 7

Math Help - Limit (two variables)

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Limit (two variables)

    Hello.

    Determine the following limit, or show it doesn't exist.

    lim [(x,y) -> (1,0)] [2(x - 1)*y^(3/2)]/[x^2 - 2x + 1 + y^3]

    WORK:

    So I tried several paths.

    Along x = 1, we have:

    lim [(1,y) -> (1,0)] [2(0)y^(3/2)]/y^3 = 0/y^3 = 0

    Along y = 0, we have:

    lim [(x,0) -> (1,0)] 0/[x^2 - 2x + 1] = 0

    So, it *appears* the limit exists so far, but in reality it exists until you can prove otherwise, that is, find a path that does not = 0. If it does exist, how do I show that it exists (and there is, in fact, NOT a path out there that does not = 0).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2006
    Posts
    221
    Okay, so apparently the limit does NOT exist, although I have yet to find a contradiction, that is, a path that does not equal 0 .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Try x=y -->0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2006
    Posts
    221
    How can you let x = y?

    You'd have:

    lim [(y,y) -> (1,0)] [2(y - 1)*y^(3/2)]/[y^2 - 2y + 1 + y^3]

    But y would approach 1 and y would approach 0? I thought they can only approach 1 value. So, you could only do it if it was (0,0) or something
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Okay I see what the problem is.

    lim (x,y)-->(1,0) of [2(x-1)y^(3/2)]/[(x-1)^2+y^3]

    You can simplify this as,

    lim (x,y)-->(0,0) of [2xy^(3/2)]/[x^2+y^3]

    Because, x-1 approaches x as x approaches 1.
    So you can replace the limit coordinates with this substitution.

    And now, try different paths.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by ThePerfectHacker View Post
    Okay I see what the problem is.

    lim (x,y)-->(1,0) of [2(x-1)y^(3/2)]/[(x-1)^2+y^3]

    You can simplify this as,

    lim (x,y)-->(0,0) of [2xy^(3/2)]/[x^2+y^3]

    Because, x-1 approaches x as x approaches 1.
    So you can replace the limit coordinates with this substitution.

    And now, try different paths.
    Thanks for the help, TPH. I never would have thought of doing this.. quite clever!

    However, I tried many paths and I still get the limit is 0..

    WORK:

    So we have lim (x,y)-->(0,0) of [2xy^(3/2)]/[x^2+y^3]

    Let x = y; then, lim (y,y) -->(0,0) of [2y*y^(3/2)]/[y^2 + y^3] = lim (y --> 0) of [2y^(5/2)]/[y^3 + y^2] = lim (y --> 0) of [2*sqrt(y)]/[y + 1] = 0..

    Similarly, if you let y = x you'll get the same thing.

    If you let x = 0, you get

    lim (0,y) --> (0,0) of [0/y^3] = 0

    And if you let y =0, you get the same thing...

    Still can't find a path where the limit is not 0.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    This one happens to be very hard, to get.
    Attached Thumbnails Attached Thumbnails Limit (two variables)-picture6.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limit (two variables)
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 7th 2012, 06:37 AM
  2. limit of several variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 18th 2011, 02:55 PM
  3. Limit with two variables
    Posted in the Calculus Forum
    Replies: 25
    Last Post: September 15th 2010, 11:45 PM
  4. Limit of functions of 2 variables
    Posted in the Calculus Forum
    Replies: 10
    Last Post: August 16th 2010, 12:40 PM
  5. Limit with Three Variables
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 8th 2009, 09:52 PM

Search Tags


/mathhelpforum @mathhelpforum