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Finding the limit for this function of two variables

  1. Oct 6, 2012 #1

    lo2

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

    Ok I have this function:

    [itex]f(x,y)=\sqrt{4xy-3y^2}[/itex]

    And then I have to find the limit:


    [itex]\lim_{h\rightarrow +0}{\frac{f(h,rh)}{h}}[/itex] Where [itex]r \in [0;\frac43][/itex]

    2. Relevant equations



    3. The attempt at a solution

    Well I must admit that I do not really understand the notation of this limit, does it mean that I have to replace x with h and y with rh, and then look at the limit? Or is it some kind of cylindric coordinate system, and if so where is the [itex]\Theta[/itex]?
     
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  3. Oct 6, 2012 #2

    tiny-tim

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    hi lo2! :smile:
    yes, that's exactly what f(h,rh) means :wink:
     
  4. Oct 6, 2012 #3

    lo2

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    Hey thanks for the help!

    I have come up with this answer:

    [itex]\lim_{h\rightarrow +0}{\frac{f(h,rh)}{h}}=[/itex] Where [itex]r \in [0;\frac43][/itex]

    [itex]\frac{\sqrt{4h^2r-3r^2h^2}}{h}=\frac{h\sqrt{4r-3r^2}}{h}=\sqrt{4r-3r^2}=
    \sqrt{r(4-3r)}[/itex]

    So this means that this limit does not depend on h really and that it only depends on the value of r. Would you deem this answer satisfactory?
     
  5. Oct 6, 2012 #4

    tiny-tim

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    yes, looks fine :smile:

    (perhaps you should mention that h > 0, so it's always h, not -h, when you take h2 outside the square-root)
     
  6. Oct 6, 2012 #5

    SammyS

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    To answer some of the questions in that last paragraph:

    Expressing (x, y) as (h, rh), means that x = h, and y = rh. That's a parametric form of y = r x, so that's a line with a slope r that passes through the origin. Therefore, this has nothing to do with cylindrical (nor polar) coordinates.

    In my opinion, that's a rather strange limit.

    [itex]\displaystyle \lim_{h\to 0^+}{\frac{f(h,rh)}{h}}[/itex] is the limit of [itex]\displaystyle \frac{f(x,y)}{x}[/itex] as (x, y) → (0, 0) along a line with slope from 0 to 4/3 .
     
  7. Oct 6, 2012 #6

    lo2

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    Ok thanks! :) Yeah I probably should do that.


    Btw if h approaches h from the left (h -> 0-) then the limit would be the negative of what I have found right?
     
  8. Oct 6, 2012 #7

    tiny-tim

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    right! :smile:
     
  9. Oct 6, 2012 #8

    lo2

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    Thanks Tiny Tim! I am glad you made it through a Christmas Carol ;)
     
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