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Math Help - L'Hospital's rule

  1. #1
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    L'Hospital's rule

    This may or not employ L'Hospital's, but here it is:


    Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

    -cos(x)-1
    --------------
    6x

    which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.
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  2. #2
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    u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true. in your case, one more time will do it. \frac{d}{dx}(-\cos(x))=\sin(x) (YOU SHOULD KNOW THAT!)
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Evan.Kimia View Post
    This may or not employ L'Hospital's, but here it is:


    Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

    -cos(x)-1
    --------------
    6x

    which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.
    Try using Maclaurin series if you know them.
    Quote Originally Posted by vince View Post
    u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true)
    Prove it!
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  4. #4
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    Ok so i tried going about it a 3rd time and get

    sin(x)
    --------
    6

    which still gives me 0 which isnt correct...

    and i dont know the Maclaurin series
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  5. #5
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    I got it. I wrote the 1st go wrong.

    Should be

    -cos(x)
    -----------
    3x^2

    then the 2nd time...
    sin(x)
    ---------
    6x

    then the 3rd time....
    -cos(x)
    ----------
    6

    which gives me -1/6 (which was right).

    Thanks though!
    Quote Originally Posted by Evan.Kimia View Post
    This may or not employ L'Hospital's, but here it is:


    Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

    -cos(x)-1
    --------------
    6x

    which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Try using Maclaurin series if you know them.


    Prove it!
    nice use of the sun...very cool.
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    Prove it!
    What?!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    What?!
    Haha, I actually thought of you when I said it.
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Haha, I actually thought of you when I said it.
    You're only human - how could you NOT think of me
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