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Homework Help: Limits of Functions of Several Variables

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f: R2 -> R be defined by: [tex]f(x,y)= \frac{4x+y-3z}{2x-5y+2z}[/tex]

    Determine if the [tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] exists. If the limit exists prove it. If not prove that it doesn't.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure what it means by "proving". I don't know if we need to show a rigorous proof (like epsilon & delta) or simply showing whether there is a common limit along different paths.

    Here is my attempt:

    As (x,y) -> (0,0) along the y-axis, x=0:

    [tex]\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{0+y-3z}{0-5y+2z}[/tex]

    As (x,y) -> (0,0) along the x-axis, y=0:

    [tex]\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+0-3z}{2x-0+2z}[/tex]

    As (x,y) -> (0,0) along the line y=x:

    [tex]\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+x-3z}{2x-x+2z} = \frac{5x-3z}{x+2z}[/tex]

    I'm a little confused here about the "z" & I don't know how to get rid of it...
    Any help is greatly appreciated.
  2. jcsd
  3. May 18, 2009 #2


    Staff: Mentor

    Your function f is a map from R3 to R, not R2 to R, so it should be written as f(x, y, z) and the limit should be as (x, y, z) --> (0, 0, 0).

    Keeping that in mind, it looks like you're on the right path. I believe that it would be sufficient to check straight-line paths into the origin; e.g y = Ax, y = Bz, x = Cz, and so on. Hope that helps.
  4. May 18, 2009 #3
    I was under the impression that checking different paths is only good for proving that the limit doesn't exist.
  5. May 18, 2009 #4
    Sorry, that was a typo but I'm not quite sure what you mean.

    As (x,y,z) -> (0,0,0) along the y-axis, x=0:

    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{y-3z}{-5y+2z}[/tex]

    As (x,y,z) -> (0,0,0) along the x-axis, y=0:

    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x-3z}{2x-2z}[/tex]

    As (x,y,z) -> (0,0,0) along the line y=x:

    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+x-3z}{2x-x+2z} = \frac{5x-3z}{x+2z}[/tex]

    It's still the same... :uhh:
  6. May 18, 2009 #5


    Staff: Mentor

    Along the y-axis, x and z are zero. Similarly, along the x-axis, y and z are zero, and along the z-axis, x and y are zero.
  7. May 18, 2009 #6


    Staff: Mentor

    Right. Before committing to proving that a limit has a certain value, it's a good idea to check that the limit actually exists. If you get different values along different paths, then you know that the limit does not exist.
  8. May 19, 2009 #7
    Thanks very much!!
    I think testing along the line y=x is only useful for the functions which are a map from R2 to R.

    So, here's my working:

    along the y-axis:

    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \lim_{(x,y,z) \to (0,0,0)} \frac{y}{-5y} = 0[/tex]

    along the x-axis:

    [tex]= \lim_{(x,y,z) \to (0,0,0)} \frac{4x}{2x} = 0[/tex]


    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{-3z}{2z} = 0[/tex]

    Hence the limit exists and equals 0.
  9. May 19, 2009 #8


    User Avatar
    Science Advisor

    NO!! How in the world did you arrive at that? For example if y= 0.000001, very close to 0, y/(-5y)= 0.000001/-0.000005= 1/5, so the limit can't be 0.

    Again, no.

    Yet again, no.

    No, it is not. Your problem is not with "limits of functions of several variables". You've done three limits of single variable functions here incorrectly.
  10. May 19, 2009 #9
    Check those limits again. They're all different.

    But even if they were all the same, you would still have to check every conceivable path to the origin to prove that the limit exists.

    Sometimes for functions of two variables you can convert to polar coordinates to prove that the limit exists. Otherwise, use the formal definition if you think you know the limit.
  11. May 19, 2009 #10

    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{y}{-5y} = \frac{1}{-5}[/tex]


    [tex]= \lim_{(x,y,z) \to (0,0,0)} \frac{4x}{2x} = 2[/tex]


    [tex]\lim_{(x,y,z) \to (0,0,0)} \frac{-3z}{2z} = \frac{-3}{2}[/tex]

    Hence the limit doesn't exists since there's no common limit. Is this enough to show that the limit doesn't exist?

    I don't know how to that...
  12. May 19, 2009 #11


    Staff: Mentor

    You now have enough information to conclude that the limit doesn't exist.
  13. May 19, 2009 #12
    Yes. All you have to show is that 2 different paths to the origin give different limits.

    You can't possibly check every path to the origin. If after you have checked a few different paths it appears that the limit exists, use the formal definition of a limit (or sometimes a change of variables) to show that the limit exists.
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