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Bounding this expression

  1. Nov 5, 2005 #1
    Hi,
    could you please help me to show this expression goes to 0 az both [itex]h_1[/itex] and [itex]h_2[/itex] go to 0 and if I know that [itex]\alpha > 3[/itex]?

    [tex]
    \left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}\right|
    [/tex]

    Thank you!
     
  2. jcsd
  3. Nov 5, 2005 #2

    HallsofIvy

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    First, remember that with two variables, it is not enough to just let one variable go to 0 and then the other. It is quite possible for a function to give two different limits if you approach (0,0) along two different paths. You need to show that if (x,y) is sufficiently close to (0,0), in any direction, then the function value will be 0.
    I think it is best to transform the function to polar coordinates so that a single variable, r, measures the distance to the origin. That's especially easy in this problem: taking [itex]h_1[/itex] on the x-axis and [itex]h_2[/itex] on the y-axis, [itex]h_1^2+ h_2^2= r^2[/itex] and [itex]h_1= r cos(\theta)[/itex].
    Put those into the formula and you will see immediately why that [itex]\alpha> 3[/itex] is necessary.
     
  4. Nov 5, 2005 #3

    benorin

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    I'll try

    [tex]\lim_{h_1,h_2\rightarrow 0} \left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}\right| = 0 [/tex]

    if and only if

    [tex] \forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{h_{1}^{2}+h_{2}^{2}}<\delta \Rightarrow \left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}} - 0\right|<\epsilon[/tex]

    Since [itex]\alpha > 3[/itex], set [itex]\alpha =3+\beta[/itex] for [itex]\beta >0[/itex].

    Choose [tex]\delta = \epsilon^{\frac{1}{\beta}}[/tex] so that if [tex]0<\sqrt{h_{1}^{2}+h_{2}^{2}}<\delta[/tex], then

    [tex]\left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}} - 0\right|=\frac{\left|h_1\right|^{3+\beta}}{(h_1^2+h_2^2)^{\frac{3}{2}}}=\frac{\left|h_1\right|^{\beta} \left|h_1\right|^3 }{(h_1^2+h_2^2)^{\frac{3}{2}}}=\frac{\left|h_1\right|^{\beta} \left(h_1^2\right)^{\frac{3}{2}} }{(h_1^2+h_2^2)^{\frac{3}{2}}} \leq \frac{\left|h_1\right|^{\beta} \left(h_1^2+h_2^2\right)^{\frac{3}{2}} }{(h_1^2+h_2^2)^{\frac{3}{2}}} = \left|h_1\right|^{\beta} = \left(h_1^2\right)^{\frac{\beta}{2}} \leq \left(h_1^2+h_2^2\right)^{\frac{\beta}{2}} [/tex]
    [tex] = \left[\left(h_1^2+h_2^2\right)^{\frac{1}{2}} \right]^{\beta} < \delta^{\beta} = \left(\epsilon^{\frac{1}{\beta}}\right)^{\beta} = \epsilon[/tex]
     
    Last edited: Nov 5, 2005
  5. Nov 5, 2005 #4

    benorin

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    It might be better (in the above proof) to replace [itex]\beta[/itex] with [itex]\alpha -3[/itex].
     
  6. Nov 5, 2005 #5
    Thank you HallsoftIvy, I tried it and is surprisingly easy to prove the limit is zero using polar expression of [itex]h_1[/itex] and [itex]h_2[/itex].

    There's just one thing I'm not completely sure about yet:
    why is it correct to put

    [tex]
    h_1^2 + h_2^2 = r^2
    [/tex]

    ? I see it is suitable to look at [itex]h_1[/itex] and [itex]h_2[/itex] as they lie on common circle with radius going zero (and thus both points going to 0), but aren't we ommiting something within two-variable functions problematics?

    I don't know whether I expressed myself in an understandable way...
     
  7. Nov 5, 2005 #6

    HallsofIvy

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    Having set up a Cartesian coordinate system with x= h1 and y= h2, then polar coordinates are DEFINED by
    [tex]r= \sqrt{h_1^2+h_2^2[/tex] and [tex]\theta= arctan(\frac{h_2}{h_1})[/tex]
    Conversely,
    [tex]h_1= r \cos(\theta)[/tex]
    Would the "something" you feel is being omitted be the angle [itex]\theta[/itex]?

    It is not omitted. Your formula
    [tex]\left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{ h_1^2+h_2^2}}\right|[/tex]
    in polar coordinates is
    [tex]\left|\frac{r^\alpha cos^\alpha(\theta)}{r^3}\right|[/tex]
    which is not only simpler but has the crucial property that if [itex]\alpha> 3[/itex], then
    [tex]\left|r^{\alpha-3} cos^\alpha(\theta)}\right|[/itex]
    goes to 0 as r goes to 0 no matter what [itex]\theta[/itex] is!
    Of course, if [itex]\alpha= 3[/itex], that does depend on [itex]\theta[/itex] and so the limit does not exist.
     
    Last edited: Nov 5, 2005
  8. Nov 5, 2005 #7
    Please could you clarify what exactly do you mean with this?

    Either:
    1) if the limit depends on [itex]\theta[/itex], it doesn't exist (not just in this case but anywhere I use polar coordinates)

    2) the limit does exist, but it is not 0 then (which is what I need to show existence of total differential)

    3) I interpreted it in a wrong way, it's another case
     
  9. Nov 5, 2005 #8
    Well if [tex] \alpha = 3 [/tex] the [tex] \frac{r^3}{r^3} [/tex] are going to cancel leaving only the [tex] \cos^3(\theta) [/tex] expression, which goes up and down, up and down, up and down ;) and somewhere at some [tex] \theta [/tex] value, the expression will actually be 0.
     
  10. Nov 5, 2005 #9
    Yes, so the right answer for [tex] \alpha = 3 [/tex] would be that the limit does exist, but depends on [tex] \theta [/tex], not that it doesn't exist as I understood it from HallsoftIvy's last response..
     
  11. Nov 5, 2005 #10

    benorin

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    But the limit as h1,h2->0 only implies that r->0, theta must be free to vary since h1 and h2 may approach the origin along any curve. If the value of theta affects the value of the limit in polar coordinates, then the limit does not exist is rectangular coordinates, as such would indicate that the limit has different values depending on how (along what curve) h1,h2->0.
     
  12. Nov 5, 2005 #11
    Thank you benorin, that's what I was asking for, now I hope I'm sure about that matter, so anytime [itex]\theta[/itex] would affect value of limit, it doesn't exist.

    This is just the same case as in rectangular coordinates - when I approach to origin with one variable directly and with the second one on straight line and the limit would depend on [itex]k[/itex] (if let's say [itex]y = kx[/itex]).
     
  13. Nov 5, 2005 #12

    benorin

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    Exactly

    Exactly.

    Yes, in the case that k effects the value of the limit, then the limit d.n.e. (does not exist), for example: the case when the value of the limit when approaching along the negative x-axis and the value of the limit when approaching along the positive y-axis are not equal, the the limit d.n.e. (e.g., if [itex]\lim_{x\rightarrow 0^{-}}f(x,0)=-3[/itex], and if [itex]\lim_{y\rightarrow 0^{+}}f(0,y)=0[/itex], then [itex]\lim_{(x,y)\rightarrow (0,0)}f(x,y)=d.n.e.[/itex].)

    This sort of thing would correspond to the case where the polar limit would vary with [itex]\theta[/itex], (and hence the limit d.n.e.); in the above example: the limit when approaching along the negative x-axis would be with [itex]\theta = \pi[/itex], and the value of the limit when approaching along the positive y-axis would be with [itex]\theta = \frac{\pi}{2}[/itex], (e.g., if [itex]\lim_{r\rightarrow 0^{+}}f(r\cos(\pi),r\sin(\pi))=-3[/itex], and if [itex]\lim_{r\rightarrow 0^{+}}f(r\cos(\frac{\pi}{2}),r\sin(\frac{\pi}{2}))=0[/itex], then [itex]\lim_{r\rightarrow 0^{+}}f(r\cos(\theta),r\sin(\theta))=d.n.e.[/itex].)

    Note that: [itex]\lim_{r\rightarrow 0^{+}}f(r\cos(\theta),r\sin(\theta))=\lim_{(x,y)\rightarrow (0,0)}f(x,y)[/itex]
     
    Last edited: Nov 5, 2005
  14. Nov 5, 2005 #13

    HallsofIvy

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    Yes, but not relevant. For some other [itex]\theta[/itex] value, the expression will be non-zero. Therefore, the limit itself does not exist.
     
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