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Math Help - Having trouble evaluating this limit

  1. #1
    Junior Member eddie2042's Avatar
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    Having trouble evaluating this limit

    y = lim [X->0] (ln x)/ ln ( cos x)

    can't use L'Hopital's rule since the indeterminant form is infinity/0. you can use it only when it's 0/0 or infinity/infinity.

    the final answer is positive infinity, but i don't know how to get there. been at it for days

    Thanks
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  2. #2
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    limit as x goes to 0 of - Wolfram|Alpha[%28Log[x]%29%2F%28Log[Cos[x]]%29]

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  3. #3
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    Quote Originally Posted by eddie2042 View Post
    y = lim [X->0] (ln x)/ ln ( cos x)

    can't use L'Hopital's rule since the indeterminant form is infinity/0. you can use it only when it's 0/0 or infinity/infinity.

    the final answer is positive infinity, but i don't know how to get there. been at it for days

    Thanks


    The limit must be \lim_{x\to 0^+}\frac{\ln x}{\ln\cos x} , since the logarithm is defined only on the positive reals, but then the numerator goes to -\infty whereas the denominator

    goes to \ln 1=0 from the left since |\cos x| \leq 1 , so it also is negative all the time.

    Tonio
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  4. #4
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    any time the denominator goes to zero or the numerator goes to infinity the expression goes to infinity

    Similarly any time the numerator goes to zero or the denominator goes to infinity the expression goes to zero

    The only exception is when both numerator and denominator approach infinity or zero
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