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Math Help - evaluating intergral

  1. #1
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    evaluating intergral

    So I have worked out this forumla

    N =  \frac{1}{3sin^2x} * (1 - cos^3x)

    As x tends to zero, N tends to 0.5, however if x =0, N = 0. Is there anyway to prove that N should be 0.5 when x = 0?

    Thanks

    Calypso
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  2. #2
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    Quote Originally Posted by calypso View Post
    So I have worked out this forumla

    N = \frac{1}{3sin^2x} * (1 - cos^3x)

    As x tends to zero, N tends to 0.5, however if x =0, N = 0.


    Says who? N isn't defined at x = 0 because you'd get \frac{0}{0} , so you can define N(0):= 0.5 .

    Tonio


    Is there anyway to prove that N should be 0.5 when x = 0?

    Thanks

    Calypso
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  3. #3
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    Quote Originally Posted by calypso View Post
    So I have worked out this forumla

    N =  \frac{1}{3sin^2x} * (1 - cos^3x)

    As x tends to zero, N tends to 0.5, however if x =0, N = 0. Is there anyway to prove that N should be 0.5 when x = 0?

    Thanks

    Calypso
    Well, strictly speaking, you can't "prove that N should be 0.5". It's not, nor is N= 0- it's undefined at x= 0. It's limit is 0.5, as x goes to 0, yes. The function, as you write it, is not continuous at x= 0. If you want to make that function continuous at x= 0, you would have to redefine it to be 0.5 at x= 0.

    Again, Tonio beat me by 1 minute!
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  4. #4
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    Is it possible to use L'hopital rule to find answer at x = 0?
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  5. #5
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    Quote Originally Posted by calypso View Post
    Is it possible to use L'hopital rule to find answer at x = 0?

    Yes. That's how I did it.

    Tonio
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    thanks
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