Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functions quiestions

  1. Oct 4, 2007 #1
    For a single variable function f(x), what does it mean when we state that lim x->3 f(x) exists?

    Also for the multivariable function f(x, y), what does it mean when we state that lim x->3, y->2 f(x, y) exists?

    Also how can I graph z=1/4((x-5)^2+(y+3)^3)^1/2 on the xy-plane as a set of contours representing static values of z.
    And how can I graph the same function but on the yz-plane as a set of contours representing static values of x.

    Thanks ^_^
  2. jcsd
  3. Oct 4, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It means that there is a number L, called the limit of f as x approaches 3 such that no matter how small a nbhd of L you look at, there is always a nbhd of of 3 such that all the neighbors of 3 are mapped in the small nbhd of L.

    This what the epsilon-delta definition is saying in words.

    Even more intuitively, it means that as you considers points nearer and nearer to 3, their image becomes nearer and nearer to L.

    It's the exact same idea.

    In both cases, you fix a value for z and you look at the relation btw x and y that pops out. For instance, fixing z=1, the equation becomes


    But, squaring both sides, this is equivalent to

    1 = 1/4((x-5)^2+(y+3)^3)

    equivalent to


    equivalent to

    1/4 - (x-5)^2 = (y+3)^3

    equivalent to


    equivalent to

    y = [1/4 - (x-5)²]^{1/3} - 3

    this is a function y(x) that you can plot relatively easily on the xy plane.
    Last edited: Oct 4, 2007
  4. Oct 4, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    Limit of f(x) as x ---> c exists if there is a number L that satisfies the following:

    For each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c| < δ, we have |f(x) − L| < ε.

    For vectors it is a similar definition where each variable has two dimensions and the absolute value operator is replaced by the norm operator ||.||.

    z = f(x,y) can be rewritten as 0 = f(x, y) - z, where z is treated as a constant. If you solve y in terms of x and z, then for each constant value of z you can express y as a contour function of x.

    On the yz plane it is the same idea except now you should treat x as a constant and solve y in terms of z (and x). Or solve z in terms of y (and x).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook