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Math Help - using a series to evaluate a limit:

  1. #1
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    using a series to evaluate a limit:

    How would I change this problem into a series and evaluate the limit?

    (e^x) - 1
    --------
    sinx

    any help greatly appreciated as I'm completely stuck.
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  2. #2
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    The MacLaurin series for [tex]e^x is 1+ x+x^2/2+ x^3/3!+ ...
    What do you get when you subtract 1 from that.

    The MacLaurin series for sin(x) is x- x^3/3!+ x^5/5!+ .... If you divide the previous series by that, do you see that you can cancel an "x"? Then, as x goes to 0, all but the constant terms will disappear.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The MacLaurin series for [tex]e^x is 1+ x+x^2/2+ x^3/3!+ ...
    What do you get when you subtract 1 from that.

    The MacLaurin series for sin(x) is x- x^3/3!+ x^5/5!+ .... If you divide the previous series by that, do you see that you can cancel an "x"? Then, as x goes to 0, all but the constant terms will disappear.
    ok, so the first x's cancel and then as I divide. (x^2/2! divided by -x^3/3! etc...)

    I wind up with:

    -3x^-1 + 20x ^-2 - 210x^ -3 etc....

    So 2 things:

    1) is my division right? there is clearly a pattern just not sure if its the right one.

    2) If it is right do I just need to take the derivative of each and calculate the limit from there or is there more i'm missing?


    edit: as x goes to 0 how would all the x's dissapear. Wouldnt I just get a bunch of 0 over 0's?
    Last edited by ryu991; May 7th 2009 at 10:08 AM.
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  4. #4
    Senior Member Spec's Avatar
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  5. #5
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    Quote Originally Posted by Spec View Post
    I'm sorry I dont understand what that means. I tried the wiki link but that just confused me more.
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  6. #6
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    Instead of messing around with all the c_2x^2+c3x^3 + ... + c_nx^n terms, you can simply say that e^x = 1+ x+O( x^2) with an error smaller than |kx^2| as x \to 0, where k is a constant. In particular, \lim_{x \to 0}O(x^n)=0 for all n > 0.
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  7. #7
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    Quote Originally Posted by Spec View Post
    Instead of messing around with all the c_2x^2+c3x^3 + ... + c_nx^n terms, you can simply say that e^x = 1+ x+O( x^2) with an error smaller than |kx^2| as x \to 0, where k is a constant. In particular, \lim_{x \to 0}O(x^n)=0 for all n > 0.
    the problem with that is my professor specifically wants us to use a series to evaluate the limit so I wouldnt be allowed to do it that way. thanks for the suggesstion though.
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  8. #8
    Senior Member Spec's Avatar
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    It's basically the same thing. I have just truncated the Maclaurin series for e^x and \sin x a bit so you don't have to write out all the terms. You can of course still write it like this:

    \lim_{x \to 0}\frac{(1 + x + \frac{x^2}{2!} + ... + \frac{x^{n}}{n!})-1}{x+\frac{x^3}{3!}+...+\frac{x^{2n+1}}{(2n+1)!}}=  \lim_{x \to 0}\frac{1 + \frac{x}{2!} + ... + \frac{x^{n-1}}{n!}}{1+\frac{x^2}{3!}+...+\frac{x^{2n}}{(2n+1)  !}}=1
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  9. #9
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    I think I got the first part, I write out all the terms in the form you did, but I'm confused as to what to do from there.

    Why does it all equal to 1?
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  10. #10
    Senior Member Spec's Avatar
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    All the x terms vanish as x goes to zero.
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  11. #11
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    Quote Originally Posted by Spec View Post
    All the x terms vanish as x goes to zero.
    1st off thanks for all the help.

    2nd, and this is the same question I asked the other guy..

    why do all the x terms vanish. As x approaches 0 wouldnt we be stuck with a bunch of 0 over 0's?

    I know the limit is 1 because I already did it with l' hospital but I just dont understand how the x's vanish and why the series does not wind up in an inderteminate form.
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  12. #12
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    No, because you have \frac{1 + ...}{1+...} where ... is a sum of x terms that goes to zero as x goes to zero. So the limit becomes \frac{1+0}{1+0}=1
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  13. #13
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    Quote Originally Posted by Spec View Post
    No, because you have \frac{1 + ...}{1+...} where ... is a sum of x terms that goes to zero as x goes to zero. So the limit becomes \frac{1+0}{1+0}=1
    ahhhh, that was so obvious I cant believe I didnt notice that. Thanks so much for all your help.
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