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Math Help - Help with a limit

  1. #1
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    Help with a limit

    Hi guys
    Could please help me with this limit:

    I am not that good with natural log and exponential functions
    Thanks in advance
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    This is good

    \ln(y)=\frac{\sin(x)}{\cos(x)}\ln(\sin(x))=\frac{1  }{\cos(x)}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}

    Now the 2nd factor is an indeterminate from so L'hospitials rule applies.

    Can you finish from here?
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  3. #3
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    could you help me more! I am kinda stuck again
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by metallica007 View Post
    could you help me more! I am kinda stuck again
    Remember that

    \lim_{x \to a}f(x)g(x)=f(a)g(a)

    \ln(y)=\frac{\sin(x)}{\cos(x)}\ln(\sin(x))=\frac{1  }{\cos(x)}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}

    So we just need to analyze the 2nd factor

    \frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}

    Using L'hosipitals rule we get

    \lim_{x \to 0}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}=\lim_{x \to 0} \frac{\frac{\cos(x)}{\sin(x)}}{\frac{-\cos(x)}{\sin^2(x)}}=\lim_{x \to 0}-\sin(x)=0

    This tells us the limit is

    \ln(y)=0 \implies y=1
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