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Math Help - limit as x approaches pi/2 from the right side using l'Hospitals's Rule

  1. #1
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    limit as x approaches pi/2 from the right side using l'Hospitals's Rule

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



    I found that:

    \lim_{x \to(\pi/2)^+}cosx=0
    \lim_{x \to(\pi/2)^+}1-sinx=0

    so I found the derivatives of them and I got:
    \lim_{x \to(\pi/2)^+}\frac{sinx}{cosx}=\frac{sin(\pi/2)}{cos(\pi/2)}=\frac{1}{0}

    I'm not sure on what I am doing wrong.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule

    lim from right -> -\infty
    Thanks from amthomasjr
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  3. #3
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    Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule

    Quote Originally Posted by amthomasjr View Post
    \lim_{x \to(\pi/2)^+}\frac{cosx}{1-sinx}



    I found that:

    \lim_{x \to(\pi/2)^+}cosx=0
    \lim_{x \to(\pi/2)^+}1-sinx=0

    so I found the derivatives of them and I got:
    \lim_{x \to(\pi/2)^+}\frac{sinx}{cosx}=\frac{sin(\pi/2)}{cos(\pi/2)}=\frac{1}{0}

    I'm not sure on what I am doing wrong.
    You haven't, technically, done anything wrong, except that, of course, because you cannot divide by 0, you cannot evaluate that final limit by simply setting x= \pi/2. What you can do now is argue that, for x close to \pi/2, but larger, sin(x) will be very close to 1 while cos(x) will be very close to 0 and negative. That gives the result that the limit is " -\infty" as MaxJasper said. I will add that " -\infty" is not a real number so that is just saying "the limit does not exist", in a particular way.
    Last edited by HallsofIvy; October 15th 2012 at 04:14 PM.
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  4. #4
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    Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule

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