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Homework Help: Limits & Continuity (Multivariable)

  1. Sep 23, 2007 #1
    I was trying to solve the practice problems in my textbook, but I am highly frustrated. The terrible thing is that my textbook has a few to no examples at all, just a bunch of theorems and definitions, so I have no idea how to solve real problems...I am feeling desperate...

    Note: Let x E R^n, f is continuous at a iff
    lim f(x) = f(a)

    1) Let f(x,y) = sin(xy) / x for x not=0. How should you define f(0,y) for y E R so as to make f a continuous function on all of R^2? Justify your reasonings.
    [I know that what I have to find is the limit of f(x,y) as (x,y)->(0,y) and set this limit value equal to f(0,y), but my trouble is that I have no idea how to evaluate limit of f(x,y) as (x,y)->(0,y)]

    2) Let f(x,y) = (xy) / (x^2 + y^2). Show that, although f is discontinuous at (0,0), f(x,a) and f(a,y) are continuous functions of x and y, respectively, for any a E R (including a = 0). We say that f is "separately continuous" in x and y.
    [f(x,a) = f(x) = (ax) / (x^2 + a^2)
    If "a" not =0, f(x) is a rational function, so it's continuous everywhere in its domain, i.e. for all x E R
    If a=0, f(x) = 0/x^2, so f(x) is discontinuous at x=0
    This is what I've done so far, and this is driving me crazy, especially in the last line where I got that for a=0, f(x) is discontinuous at x=0. So I must have done something wrong, but I have no idea where...]

    3) Let f(x,y) = y(y - x^2) / x^4 if 0 < y < x^2, and f(x,y)=0 otherwise. Find ALL point(s) for which f is discontinuous.
    [This looks really hard. How can I approach this problem? Any help?]

    Any help is greatly appreciated!
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 24, 2007 #2
    Can someone please help me?
  4. Sep 24, 2007 #3
    hm..for #1 i think when you have (x,y)->(0,y) you set x=0 and then evaluate the limit as y->0.

    for #2 if x^2 can go to 0 then yea it's discontinuous but if x^2 can't equal 0 then f(x)=0 which would leave it continuous, but it doesn't say...

    for #3 what points would you expect to not be discontinous? from the wording of the problem it states alot of points you could suspect to not be continous since the limit of f(a,b) won't equal f(a,b)
  5. Sep 24, 2007 #4
    1) But when I set x=0, f(0,y)= 1 / 0, now how can I evaluate the limit of f(0,y) as y->0 ?

    2) So is there any way to find out whether for a=0, f(x) is discontinuous at x=0 or not?

    3) Would the "possible" discontinuous points be y=x^2? Do I just have to check the points along this curve?
  6. Sep 24, 2007 #5
    1) hm...you would need to somehow separate x and y to be able to make x=0 and leave some function of y...

    2)does the domain of the function include (0,0)? if no then x,y cannot go to (0,0) so 0/x^2 will equal?

    3)hm..maybe but the function and the limit is going to be different for each point when 0<y<x^2 so x^2 seems unlikely.
    You're looking for any points where the lim f(x,y) as (x,y)->(a,b) =f(a,b). if any points don't pass that test then the function is discontinous there.
  7. Sep 24, 2007 #6
    1) How can I separate x and y? I have no idea...

    2) The domain of f(x,y) doesn't include (0,0).
    For a=0, f(x,a) = f(x) = 0/x^2
    But we are trying to prove that "f(x,a) is continuous function of x", this means that we have to prove that "f(x,a) is continuous function of x for all x E R", right? By the word continuous alone, without specifying the intervals, I would understand it as being continuous everywhere.

    3) But before checking, how can I even locate all the possible candidates for which a discontinuity may occur?
  8. Sep 24, 2007 #7
    1) duno either >< was thinking series expansion? but hm....

    2)well if x ever = 0 then it becomes discontinous, i think the problem does mean all x in the domain of the function which would b R\{0} and if a=0 you already know the limit so that would make it continous. and the same for y. But if it means ALL x E R even x=0 then.....I don't think there's a way...

    3)in order for a function to be continous at a point (a,b):
    (a,b) E Domain of the function
    (a,b) is not an isolated point
    lim f(x,y) as (x,y)->(a,b) exists and = L

    so are there any points that you can see from the problem that could fail any of those? Usually you encouter discontinuities with the limits.
  9. Sep 25, 2007 #8
    2) But if a funciton f is discontinuous at x=0, would you still call f a continuous funciton?

    3) I am having terrible trouble with continuity here since it's a function of 2 variable. My understandings for continuity are perfect for functions with one variable, but I have absolutely no idea what is going on for the function here. For example, how can I know if f(x,y) is continuous at the point (0,1) or not?

    Thanks a lot!
  10. Sep 25, 2007 #9
    2) hm...i don't know really the wording is vague ><...

    3) well at point (0,1) first you find whether you use the top or bottom definition.
    0<y<x^2 for the top and all other y's for the bottom.
    x=0, 0<1<0^2 = 0<1<0 which isn't true so you use the bottom formula:
    so f(0,1) = 0. if you compute the limit of f(x,y) you will use the top function which will not have a limit so the function will be discontinous at (0,1).
    but you want to find all points where it's not discontinous. I recommend looking any any points where 0<y<x^2 doesn't hold since f(x,y)=0 for all of those other numbers the limit f(x,y) won't equal 0 at all of those points so alot of them will be discontinous.
  11. Sep 27, 2007 #10
    3) "...so the function will be discontinous at (0,1)."<----But the answer in my book is that it is discontinuous ONLY at (0,0), how come?
  12. Sep 27, 2007 #11
    Are the possible candidates of discontinuity y=x^2 and y=0 ONLY? How can I know for sure the function is continuous at all other locations?
  13. Sep 27, 2007 #12


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    You can "cheat" a little bit on this one. If there is a value of of f(0,y) such that assigning f that value will make it continuous then the limit, as x goes to 0 must exist. Can you find that limit? Remember that sin(x)/x goes to 1 as x goes to 0 and think about multiplying both numerator and denominator of sin(xy)/x by y.

    ?? What? If a= 0, f(x)= 0/x^2= 0, for all x. Are you saying a constant function is discontinuous?

    This function is defined for all x,y, so the only way it could be discontinuous is if the limit does not exist or is different from the value. The only place a rational function (which this is) does MAY not have a limit is where the denominator is 0. Where is that? What is the limit at (0,y)? (Hint: the denominator has a higher power than the numerator.) What is the value of the function there?
    Since y(y-x^2)/x^4 is continuous everywhere except possibly x= 0 so the only other possible discontinuities are where the "formula" changes: on the line y= 0 and the curves y= x^2. What are the limits as you approach any point on either of those?

    Last edited by a moderator: Sep 27, 2007
  14. Sep 27, 2007 #13
    it's been a while since my multivariable class but even then I remember my teacher saying it wasn't enough to approach one way or another. I suggest changing variables: this is the same in problem 1 as multiplying by 1=y/y as halls of ivy suggested.

    f(x) is continous at the origin like halls of ivy says. Approach origin from y=x to show f(x,y) is discontinuous there.
  15. Sep 28, 2007 #14
    But f(x) is undefined (hence discontinuous) at 0, division by 0 is not allowed. Isn't it a constant function with a "hole" at x=0?

    I sort of get the idea of locating the possible discontinuities, but what limits do I have to find?
    lim f(x,y) ??
    But (0,y) is not a single point; y can be any real number...is this ok? Shouldn't I be checking the continuity of each point?
    What other limits do I need?

    f is continuous at a iff
    lim f(x) = f(a)
    This is continuity at a "point"

  16. Sep 28, 2007 #15
    it is true that division by zero isn't allowed, but in some cases since the limit as x approaches the origin (we're talkn about f(x) here btw) is finite. It's called a removable discontinuity, you can just redefine the function by placing the limit where the hole is ie you can REMOVE the discontinuity without changing any essential part of the function. A function with a removable discontinuity is basically continuous.
  17. Sep 30, 2007 #16
    3) How can I show that the 2 functions f(x,y) = y(y - x^2) and f(x,y)=0 agree completely along the boundaries y=x^2 and y=0 except at x=0?
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