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Finding a limit

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to find the following limit.


    2. Relevant equations
    [tex]\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)}[/tex]


    3. The attempt at a solution
    I think it's got to be something with Taylor series, but I don't really know how to do it.
     
  2. jcsd
  3. Nov 15, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi squeetox! Welcome to PF! :smile:

    Hint: for example, the Taylor series for e3x is 1 + 3x + … , so if (e3x - 1) was a factor, you could replace it (as x -> 0) by 3x. :wink:
     
  4. Nov 15, 2008 #3
    Thanks!


    Why do the others terms (9x2/2 + 9x3/2...) disappear as x->0?

    PS: I had thought of replacing [itex]\sin^3 x (e^{x^2}) \sim x^5[/itex] because I know that [itex]\sin [f(x)] \sim f(x)[/itex] and [itex]e^{f(x)} -1 \sim f(x)[/itex]. The rest I have no clue though.
     
    Last edited: Nov 15, 2008
  5. Nov 16, 2008 #4

    tiny-tim

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    Hi squeetox! :smile:
    They don't actually disappear … but, compared with x, they get so small (as x -> 0) that we can ignore them. :smile:
    Nooo … you're confusing ex2 and ex2 - 1. :wink:
     
  6. Nov 16, 2008 #5
    So basically you mean that since it's a sum of x's to different powers, and the ones after 3x get smaller much faster, I can ignore them?
    Thanks.
     
  7. Nov 16, 2008 #6
    I'm sorry that's what I meant. If you look at the limit I'm looking for, it's [itex]e^{x^2} - 1 \sim x^2[/itex] therefore [itex]sin^3 x(e^{x^2}-1) \sim x^5[/itex].

    Now what about [itex](x-\sinh x)(\cosh x- \cos x)[/itex] and [itex]5+\sin x \ln x[/itex]?
     
  8. Nov 16, 2008 #7

    tiny-tim

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    ah yes, so it is … i didn't look back up to the top! :redface:
    The first lot, you do the same way.

    The sinx lnx … I can't remember whether there's a Taylor expansion for lnx … it obviously does -> 0 … just look at sin10-100 ln10-100, but how to prove it? :rolleyes:

    anyway, that's you problem, not mine! :wink: :biggrin:
     
  9. Nov 16, 2008 #8
    For sin x ln x, I'm thinking...
    [tex]\sin x \ln x \sim x \ln x[/tex]

    And
    [tex]\lim_{x\rightarrow0}x \ln x = \lim_{x\rightarrow\infty} \frac{1}{x}\ln\frac{1}{x} = \lim_{x\rightarrow\infty} -\frac{\ln x}{x} = 0[/tex]

    Therefore:
    [tex]\lim_{x\rightarrow0}(5+\sin x \ln x) = 5[/tex]


    For the first lot, do I find Taylor expressions for (x - sinh x) and (cosh x - cos x) or for the product of both?

    Edit: I got it:
    [tex]x-\sinh x \sim -\frac{x^3}{6}[/tex]
    [tex]\cosh x - \cos x \sim x^2[/tex]

    So finally...
    [tex]\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)} = -\frac{1}{30}[/tex]

    Thank you so much for your help!
     
    Last edited: Nov 16, 2008
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