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Lim{x->0} function

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    [tex]\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}=\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}*\frac{\sqrt{x+b^2}+b}{\sqrt{x+b^2}+b}=\lim_{x \rightarrow 0}=\frac{(\sqrt{x+a^2}-a)(\sqrt{x+b^2}+b)}{x}[/tex]
    On the next step I divided by x, but again nothing.
    I tried several different methods by substituting [itex]y=\sqrt{x+a^2}[/itex] and [itex]z=\sqrt{x+b^2}[/itex] but useless. Please help! Thanks in advance.
     
  2. jcsd
  3. Nov 19, 2008 #2

    tiny-tim

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    Hi Дьявол! :smile:

    If you're allowed to, use l'Hôpital's rule …

    if you're not, use a Taylor expansion. :wink:
     
  4. Nov 19, 2008 #3
    @tiny-tim I have never used both of them. Aren't there any elementary transformations?
     
  5. Nov 19, 2008 #4

    tiny-tim

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    ok … here's a trick …

    whenever I see a √ sign, I think :rolleyes: of using a trig substitution, so …

    define θ and φ by x = asinθ = bsinφ, and let θ and φ tend to zero, and use standard trigonometric identities …

    what do you get? :smile:
     
  6. Nov 19, 2008 #5
    Maybe, that is nice trick. If θ and φ tend to 0 then asin0 and bsin0 tend to 0 so that x tend to 0.

    But after substituting I get the things messed up :smile:

    Should I try like this:

    [tex]\frac{\sqrt{asin(\theta)+a^2}-a}{\frac{\sqrt{asin(\theta)+\frac{a^2sin^2\theta}{sin^2(\varphi)}}}{-\frac{asin(\theta)}{sin(\varphi)}}[/tex]
    or maybe:
    [tex]\frac{\sqrt{asin(\theta)+a^2}-a}{\sqrt{bsin(\varphi)+b^2}-b}[/tex]
     
    Last edited: Nov 19, 2008
  7. Nov 19, 2008 #6

    tiny-tim

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    oops!

    Sorry … I couldn't read my own handwriting :redface:

    I should have said x = a2tan2θ = b2tan2φ …

    or even better x = a2cosθ = b2cosφ, with θ and φ -> π/2 :smile:
     
  8. Nov 19, 2008 #7
    Re: oops!

    Maybe it is better if we use x = -a2cosθ = -b2cosφ (it is the same thing), and I get:
    [tex]\frac{asin^2(\theta)-a}{bsin^2(\varphi)-b}[/tex]
    for θ and φ -> π/2
    but I can't do anything out of here, or I can (circular movement) :smile:

    If I do like this:
    -a(1-sin2θ/-b(1-sin2φ)=acos2θ/bcos2φ=0/0 :smile:
     
  9. Nov 19, 2008 #8

    tiny-tim

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    No, you get (a/b)√(1 - cosθ)/√(1 - cosφ) :smile:
     
  10. Nov 19, 2008 #9

    Avodyne

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    Well, this is all getting very complicated. It's better to learn a more universal method.

    The point is, what can we use to approximate [itex]\sqrt{x+a^2}[/itex] when [itex]x[/itex] is small?

    In general, answering this sort of question involves using a Taylor expansion,
    [tex]f(x)=f(0)+f'(0)x+{\textstyle{1\over2}}f''(0)x^2+\ldots[/tex]
    Typically only the first two terms are needed for a limit problem.

    In the case of interest, [itex]f(x)=(x+a^2)^{1/2}[/itex], and so [itex]f'(x)=\frac12 (x+a^2)^{-1/2}[/itex]. Thus we have [itex]f(0)=a[/itex] and [itex]f'(0)=1/(2a)[/itex], and so [itex]\sqrt{x+a^2}\simeq a + x/(2a)[/itex] when [itex]x[/itex] is small.
     
  11. Nov 19, 2008 #10

    Avodyne

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    Here's an alternative method for this particular problem that doesn't use Taylor expansions. Multiply the original expression by
    [tex]\Biggl(\frac{\sqrt{x+a^2}+a}{2a}\Biggr)\Biggl(\frac{2b}{\sqrt{x+b^2}+b}\Biggr)[/tex]
    which has a limit of 1 as [itex]x\to0[/itex].
     
    Last edited: Nov 19, 2008
  12. Nov 19, 2008 #11

    tiny-tim

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    ooh, that's much better!

    nice one, Avodyne! :smile:
     
  13. Nov 19, 2008 #12
    @Avodyne
    That's great. I didn't think that I could use something like that, but that's obviously lim1=1; so again I can use it in my particular problem. Thanks a lot.

    @tiny-tim
    Thanks for the replies. But let's solve it also on your way of thinking.
    If x = a2cosθ = b2cosφ, with θ and φ -> π/2
    then in the original equation:
    √(a2cosθ+a2)-a/ √(b2cosφ+b2)-b=
    =a√(1+cosθ)-a/ b√(1+cosφ)-b =a(√(1+cosθ)-1)/b(√(1+cosφ)-1)
    What could I do know? Again stuck.

    Using Avodyne's method I got b/a which is actual correct, but also I found your way very good and I would like to solve it. Thanks again.
     
  14. Nov 19, 2008 #13

    tiny-tim

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    Well. it's not pretty :blushing:

    a(√(1+cosθ)-1)/b(√(1+cosφ)-1)

    = a((√2)cosθ/2 - 1)/b((√2)cosφ/2 - 1)

    = a(cosθ/2 - 1/√2)/b((cosφ/2 - 1/√2)

    = a(cosθ/2 - cos45º)/b((cosφ/2 - cos45º)

    = a sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) / b sin(φ/4 + 22.5º) sin(φ/4 - 22.5º)

    ~ a (θ/4 + 22.5º) (θ/4 - 22.5º) / b (φ/4 + 22.5º) (φ/4 - 22.5º)

    = a (θ + 90º) (θ - 90º) / b (φ + 90º) (φ - 90º);

    using (θ + 90º)/(φ + 90º) ~ 180º/180º = 1

    and b2/a2 = cosθ/cosφ = sin(θ - 90º)/sin(φ - 90º) ~ (θ - 90º)/(φ - 90º),​

    ~ (a/b)(b2/a2) = b/a :redface: :redface:
     
  15. Nov 19, 2008 #14
    :smile:
    How do you got equal (cosθ/2 - cos45º) = sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) ?
     
  16. Nov 19, 2008 #15

    tiny-tim

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    Standard trig formula: cosA - cosB = 2 sin(A+B)/2 sin(A-B)/2 …

    work it out! :wink:
     
  17. Nov 19, 2008 #16

    Mark44

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    There's another approach that I didn't see in this thread, that's much simpler than most of the approaches that were suggested:
    Multiply the original expression by 1, in the form of
    [tex]\frac{\sqrt{x + a^2} + a}{\sqrt{x + b^2} + b} * \frac{\sqrt{x + b^2} + b}{\sqrt{x + a^2} + a}[/tex]
    This makes the limit very easy to evaluate, which when simplified, is b/a.
     
  18. Nov 19, 2008 #17

    tiny-tim

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    Hi Mark! :smile:

    Yes, that's nice …

    but Avodyne beat you to it in post #10,

    and his version was neater 'cos it had less square-roots! :wink:
     
  19. Nov 19, 2008 #18
    = a sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) / b sin(φ/4 + 22.5º) sin(φ/4 - 22.5º)

    ~ a (θ/4 + 22.5º) (θ/4 - 22.5º) / b (φ/4 + 22.5º) (φ/4 - 22.5º)

    Sorry for asking, but where the sine gone here?
     
  20. Nov 19, 2008 #19

    Mark44

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    I noticed Avodyne's approach and thought that mine was similar to his (hers?), but still different. Avodyne multiplied by something that is 1 in the limit. I mulitplied by something that is 1 (taking domain into account).
     
  21. Nov 19, 2008 #20

    tiny-tim

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    Magic! :biggrin:

    hmm … on reflection :redface:, I can only magic away the minus brackets … I have to treat the plus brackets separately …

    sin(θ/4 + 22.5º) / sin(φ/4 + 22.5º) ~ sin45º / sin45º = 1,

    and

    sin(θ/4 - 22.5º) / sin(φ/4 - 22.5º) ~ (θ/4 - 22.5º) / (φ/4 - 22.5º)

    because as θ -> 90º, (θ/4 - 22.5º) -> 0, and sinx ~ x as x -> 0 :smile:
    ah, yes, but that's what I prefer about his method … it's more compact! :wink:
     
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