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Math Help - Trig Limit Difficulty

  1. #1
    Junior Member beebe's Avatar
    Joined
    Aug 2011
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    Trig Limit Difficulty

    \lim_{x \to (\pi/2)^+} \frac{cos(x)}{1-sin(x)}

    I get: = \frac{0}{0} I attempt to apply L'Hospital's rule:

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

    L'Hospital's rule is no bueno. So, my next thought was to use the squeeze theorem, but since the numerator and denominator have different vertical boundaries, (and the function shoots off unbounded near \frac{\pi}{2}) I can't think of a way to do that. I know intuitively that the limit should be -\infty, I just can't get there analytically.
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  2. #2
    Senior Member
    Joined
    Nov 2010
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    Clarksville, ARk
    Posts
    398

    Re: Trig Limit Difficulty

    Quote Originally Posted by beebe View Post
    \lim_{x \to (\pi/2)^+} \frac{cos(x)}{1-sin(x)}

    I get: = \frac{0}{0} I attempt to apply L'Hospital's rule:

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

    L'Hospital's rule is no bueno. So, my next thought was to use the squeeze theorem, but since the numerator and denominator have different vertical boundaries, (and the function shoots off unbounded near \frac{\pi}{2}) I can't think of a way to do that. I know intuitively that the limit should be -\infty, I just can't get there analytically.
    Actually, L'H˘pital is bueno . \lim_{x \to (\pi/2)^+} \frac{-\sin(x)}{-\cos(x)}= \lim_{x \to (\pi/2)^+} \frac{\sin(x)}{\cos(x)}\ \to\ \frac{1}{0^-}\ \to\ -\infty

    You can alternatively multiply by the initial rational expression by \frac{1+\sin(x)}{1+\sin(x)} to get

     \frac{\cos(x)}{1-\sin(x)}\frac{1+\sin(x)}{1+\sin(x)}=\frac{\cos(x)(  1+\sin(x))}{1-\sin^2(x)}=\frac{\cos(x)(1+\sin(x))}{\cos^2(x)}
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