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Math Help - Help with limit

  1. #1
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    Help with limit

    Can anyone show how to solve this?

    \displaystyle\lim_{x\to\{0^+}\frac{ln x}{csc x}

    (Sorry for that weird bracket in the limit... it's just supposed to be "the limit as x goes to 0 from the right.")

    Pretty sure I'm supposed to use L'Hopital's rule for this, but when I do, I get stuck after the first derivative. Any advice?
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  2. #2
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    It will be (After using L`Hospital`s Rule): \frac{-1}{xcsc(x)\cot(x)}
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  3. #3
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    When I use L'Hopital's rule I end up with

    \frac{1/x}{-cot(x)csc(x)}
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  4. #4
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    Quote Originally Posted by paupsers View Post
    When I use L'Hopital's rule I end up with

    \frac{1/x}{-cot(x)csc(x)}
    Sorry I edited it.
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  5. #5
    Junior Member doomgaze's Avatar
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    Quote Originally Posted by paupsers View Post
    When I use L'Hopital's rule I end up with

    \frac{1/x}{-cot(x)csc(x)}
    That is correct, take the derivative again.

    probably easier to think of 1/x as x^-1
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  6. #6
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    Apply L`Hospital`s Rule again
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  7. #7
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    Using the rule again, I end up with

    \frac{-x^-2}{csc(x)(csc^2(x)-cot(x))}

    Plugging in 0 gives me \frac{0}{\infty(\infty-\infty)}

    Does that equate to 0?
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  8. #8
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    infnity - infinity is indeterminated
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  9. #9
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    So do I have to use the rule again?
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  10. #10
    Math Engineering Student
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    Don't make it too complicated:

    \sin (x)\ln (x)=\frac{\sin x}{x}\cdot x\ln x, for x\ne0, now take limits, what do you get?
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  11. #11
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    Quote Originally Posted by paupsers View Post
    So do I have to use the rule again?
    Check your derivative for the denominator.
    Are you sure your signs are correct?
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