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Continuity of a function.

  1. Dec 11, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Been awhile since I looked at this, just seeing if I still know what I'm doing here.

    Suppose : [itex]f(x,y) = \frac{x^{2/3}y^2}{x^2 + |y|^3} [/itex] for (x,y) ≠ (0,0).

    1. Show that on every straight line through the origin the limit as (x,y) → (0,0) of f(x,y) exists.
    2. Does the general limit as (x,y) → (0,0) exist? Justify!

    2. Relevant equations

    I'm thinking I might need a formal epsilon delta proof for part 2.

    3. The attempt at a solution

    1. So every straight line through the origin means that I want to consider y = mx.

    [itex]f(x, mx) = \frac{m^2x^{8/3}}{x^2 + |mx|^3}[/itex]

    Now, I have sadly forgotten what I must do here. I know I want to take the limit as x goes to zero, but my brain is having a hiccup for some reason doing the algebra. Should I divide through by x3 like I'm thinking or?

    2. I'm quite certain that this limit is not going to exist in general. Not sure how to go abouts it yet though.

    Thanks for any help in advance :).
     
  2. jcsd
  3. Dec 11, 2012 #2

    SammyS

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    Divide numerator & denominator by x2/3 .

    Right. The limit does not exist.

    A common approach is to change to polar coordinates. Otherwise, try to find a path going through the origin along which limit f(x,y) ≠ 0 as (x,y) → (0,0) . If you can find a path through the origin on which f(x,y) is a non-zero constant (except at the origin) then that does it.
     
  4. Dec 11, 2012 #3

    pasmith

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    [tex]x^2 + |mx|^3 = x^2 + |m|^3|x|^3 = x^2(1 + |m|^3|x|)[/tex]

    Cancel the [itex]x^2[/itex] factor on top and bottom and take the limit. That gets you every line through the origin except the line [itex](0,y)[/itex], which you need to investigate separately.

    EDIT: If you want one that certainly doesn't have a general limit, look at
    [tex]f(x,y) = \frac{yx}{x^2 + |y|^3}[/tex].
     
    Last edited: Dec 11, 2012
  5. Dec 11, 2012 #4

    Zondrina

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    Ahh so for part 1 :

    f(x,mx) → 0 as x → 0
    f(0,y) → 0 as y → 0
    f(x,0) → 0 as x → 0

    Therefore f(x,y) → 0 as (x,y) → (0,0).

    For part 2, I consider |f(rcosθ, rsinθ)| ( I'll skip some of the algebra to get to the more important stuff )

    [itex]|f(rcosθ, rsinθ)| = r^{2/3} |\frac{cos^{2/3}θsin^2θ}{cos^2θ + r|sin^3θ|}|[/itex]

    Hmm as r → 0, r2/3 → 0. That's not supposed to happen? Did I miss something?

    EDIT : If I pull r out of the denominator, I get that it tends to infinity as r → 0 which would mean that f(x,y) is unbounded near the origin and thus the general limit does not exist.

    Still a bit confused here though.
     
    Last edited: Dec 11, 2012
  6. Dec 11, 2012 #5

    pasmith

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    If you pull r out of the denominator you get
    [tex]
    \frac{r^{-1/3}\cos^{2/3}\theta \sin^2\theta}{\frac{\cos^2\theta}{r} + |\sin^3\theta|}
    [/tex]
    so that numerator and denominator both tend to infinity as [itex]r \to 0[/itex].

    You were right the first time: [itex]f(x,y) \to 0[/itex] from any direction as [itex]\|(x,y)\| \to 0[/itex], so the general limit exists.
     
  7. Dec 11, 2012 #6

    SammyS

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    No. f(x,y) → 0 as (x,y) → (0,0) along any straight line passing through the origin.

    First, your 'Edit comment': If you pull r out of the denominator, then the denominator goes to ∞ as r → 0. That won't get you anywhere.

    What I did was look for a path along some constant value of f(x,y). I noticed that f(1,1) = 1/2, so I solved f(x,y) = 1/2 for y . That indeed gives a path through the origin.
     
  8. Dec 11, 2012 #7

    Zondrina

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    Sorry for being so hasty. So yes, I meant f(x,y) → 0 as (x,y) → (0,0) along any straight line passing through the origin for part 1.

    For part 2, the general limit exists then since if r → 0, |f(rcosθ,rsinθ)| → 0?

    @ Sammy, how would you notice such a thing? In an exam situation I wouldn't be plugging in points frantically trying to find a path like you did. ( I'm presuming the function I'm going to be given is a bit more complicated ).
     
  9. Dec 11, 2012 #8

    HallsofIvy

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    By the way, NOT every line through the origin can be written in the form "y= mx". Any non-vertical line can be but the vertical line, x= 0, has to be done separately.
     
  10. Dec 11, 2012 #9

    SammyS

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    Look at [itex]\displaystyle \ \ f(r\cos(\theta,r\sin(\theta))=r^{2/3}\frac{(\cos^2(\theta))^{1/3}\sin^2(\theta)}{\sin^2(\theta)|\cos^3(\theta)|}\ \ [/itex] again. As r → 0 and θ → π/2, you have the form 0∙∞ . That's a problem.


    As for how you would notice that there is a path:

    Well, what I initially did won't help much on an exam, unless you can plot contour graphs. I used WolframAlpha to graph f(x.y) . It's clear from that graph that there is a problem at the origin. http://www.wolframalpha.com/input/?i=%28|x|^{2%2F3}y^2%29%2F%28x^2+%2B+|y|^3%29

    attachment.php?attachmentid=53845&stc=1&d=1355275914.gif attachment.php?attachmentid=53846&stc=1&d=1355275914.gif

    These graphs do indicate that contour lines pass through the origin.

    After playing around with this for quite a while, I conclude that I just got lucky.

    The contours do suggest [itex]\displaystyle y = x^\alpha\,,\ \text{ where }0<\alpha<1\ .[/itex]

    If you plug that into f(x,y), then the only value of [itex]\ \alpha\ [/itex] for which the exponent of the denominator is greater than or equal to the numerator is [itex]\ \alpha=2/3\,,\ [/itex] and in that case the exponents are equal.

    So, taking a path, [itex]\displaystyle y=C\,x^{2/3}[/itex] gives a non-zero limit for f(x,y) as (x,y) → (0,0) .
     

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  11. Dec 11, 2012 #10

    Zondrina

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    Ahh this is much more clear to me now. So any multiple of the path x2/3 will produce a non-zero limit which will not comply with what we found earlier along the straight lines. Hence no general limit will exist.

    So a sort of trick is to try and spot if I can find a path which will make the degree of the numerator and denominator the same which would yield a non-zero result for cases like these.

    Per say if I switched to polars and found the limit was zero then, I would conclude that the limit did exist and was zero.

    It all depends I suppose, but thank you very much for clarifying this for me Sammy. Thumbs up.
     
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