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Math Help - Limit calculation

  1. #1
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    Limit calculation

    I can't seem to get the hang of calculating these indeterminate forms...how do I figure out

    limit ln(cosx) / ln(cos5x) as x goes to pi/2 from the left

    The left limits of the numerator and denominator at pi/2 are negative infinity. So I tried looking at the limit of their derivatives: tanx / 5tan(5x) taking out the 1/5 I'm left to calculate the limit of tanx / tan5x. Once again the left limit at pi/2 is infinity.
    So looking again at the limit of the derivatives it comes out (secx)^2 / 5(sec5x)^2
    which once again I have no idea how to calculate.

    I'm very confused.
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  2. #2
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    Quote Originally Posted by moses View Post
    I can't seem to get the hang of calculating these indeterminate forms...how do I figure out

    limit ln(cosx) / ln(cos5x) as x goes to pi/2 from the left

    The left limits of the numerator and denominator at pi/2 are negative infinity. So I tried looking at the limit of their derivatives: tanx / 5tan(5x) taking out the 1/5 I'm left to calculate the limit of tanx / tan5x. Once again the left limit at pi/2 is infinity.
    So looking again at the limit of the derivatives it comes out (secx)^2 / 5(sec5x)^2
    which once again I have no idea how to calculate.

    I'm very confused.
    Hint:

    \frac{d}{dx} \ln(cos(x))=\frac{-\sin(x)}{\cos(x)}=-\tan(x)
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  3. #3
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    Okay, so the derivatives of ln(cos(x)) and ln(cos(5x)) are -tan(x) and -5tan(5x). But how do I calculate the limit -tan(x) / -5tan(5x) = (1/5) * (tan(x) / tan(5x))
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  4. #4
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    Quote Originally Posted by moses View Post
    Okay, so the derivatives of ln(cos(x)) and ln(cos(5x)) are -tan(x) and -5tan(5x). But how do I calculate the limit -tan(x) / -5tan(5x) = (1/5) * (tan(x) / tan(5x))
    \lim_{x \to 0}\frac{\tan(x)}{5\tan(5x)}=\lim_{x \to 0}\frac{1}{5}\frac{\sin(x)}{\cos(x)}\cdot \frac{\cos(5x)}{\sin(5x)}=\lim_{x \to 0}\frac{1}{5}\frac{\sin(x)}{\sin(5x)}\cdot \frac{\cos(5x)}{\cos(x)}

    \lim_{x \to 0}\frac{1}{25}\frac{\sin(x)}{x}\cdot \frac{5x}{\sin(5x)}\cdot \frac{\cos(5x)}{\cos(x)}

    Can you finish from here?
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  5. #5
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    Yes, I get it now. I wouldn't have thought of splitting up the sinx/5sin5x like that...
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