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Math Help - l'hopitals rule problem

  1. #1
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    l'hopitals rule problem

    can someone solve


    lim (sin4x)/(x^2+8x)
    x approaches 0
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by cummings15 View Post
    can someone solve


    lim (sin4x)/(x^2+8x)
    x approaches 0
    \lim_{x\to 0}\frac{\sin (4x)}{x^2+8}=\lim_{x\to 0}\frac{\cos(4x)\cdot 4}{2x+8}=\frac{4}{8}=\frac{1}{2}
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  3. #3
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    A little more basic method:

    cummings15, rewrite it as \frac{sin(4x)}{x^2+ 8x}= \frac{sin(4x)}{x(x+ 8)} = 4\frac{sin(4x)}{4x}\frac{1}{x+ 8}.

    Now, if you know the \lim_{\theta\to 0}\frac{sin(\theta)}{\theta}, this is easy.
    Last edited by HallsofIvy; May 6th 2010 at 08:01 AM.
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  4. #4
    Member grgrsanjay's Avatar
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    Unhappy :(

    wats actually a l'hopital rule
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  5. #5
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    Yes, I just noticed the title of the thread- but I still think the more fundamental method is better.
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  6. #6
    Member grgrsanjay's Avatar
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    yea got it

    l'hospital is a rule to differentiate numerator and denominator separately and then substitute the limit

    can we use it for 1/0 cases or infinity/0 cases??
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