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Math Help - Finding Limit With L'Hospital's Rule

  1. #1
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    Finding Limit With L'Hospital's Rule

    lim x->1 (2-x)^(tan(pi(x)/2))

    I plugged it in and got 1^infinity (indeterminate form) because tan (pi(1)/2) is 1/0 or infinity

    so:

    lny = (tan(pi(x)/2)) * ln(2-x)

    = ln(2-x) / cot (pi(x)/2)

    L'H = (-1/2-x) / csc^2(pi(x)/2) ( -pi/2)

    = -1/ -pi/2
    = 2/pi

    lim x->1 y = e^(2/pi)

    ?

    Where did I go wrong?

    Thanks
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  2. #2
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    Re: Finding Limit With L'Hospital's Rule

    Hello, NeoSonata!

    Why do you think your answer is wrong?
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  3. #3
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    Re: Finding Limit With L'Hospital's Rule

    I checked the answer using wolfram alpha

    it got:
    lim x->1 (2-x)^(tan(pi(x)/2)) = 1

    the site does not show any work for this problem though.
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  4. #4
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    Re: Finding Limit With L'Hospital's Rule

    Quote Originally Posted by NeoSonata View Post
    lim x->1 (2-x)^(tan(pi(x)/2))

    I plugged it in and got 1^infinity (indeterminate form) because tan (pi(1)/2) is 1/0 or infinity

    so:

    lny = (tan(pi(x)/2)) * ln(2-x)

    = ln(2-x) / cot (pi(x)/2)

    L'H = (-1/2-x) / csc^2(pi(x)/2) ( -pi/2)

    = -1/ -pi/2
    = 2/pi

    lim x->1 y = e^(2/pi)

    ?

    Where did I go wrong?

    Thanks
    You must have entered it into Wolfram wrongly, since the limit is \displaystyle e^{\frac{2}{\pi}}.

    limit of [(2-x)^(Tan[Pi * x / 2])] as x to 1 - Wolfram|Alpha

    \displaystyle \begin{align*} \lim_{ x \to 1} (2 - x)^{\tan{\frac{\pi x}{2}}} &= \lim_{x \to 1}e^{\ln{\left[(2 - x)^{\tan{\frac{\pi x}{2}}}\right]}} \\ &= \lim_{x \to 1}e^{\tan{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}} \\ &= e^{\lim_{x \to 1}\tan{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}} \\ &= e^{\lim_{x \to 1}\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}}{\cos{\left(\frac{\pi x}{2}\right)}}} \\ &= e^{\lim_{x \to 1}\frac{\frac{\pi}{2}\cos{\left(\frac{\pi x}{2}\right)}\ln{(2-x)} - \frac{\sin{\left(\frac{\pi x}{2}\right)}}{2-x}}{-\frac{\pi}{2}\sin{\left(\frac{\pi x}{2}\right)}}}\textrm{ by L'Hospital's Rule} \\ &= e^{\frac{\frac{\pi}{2} \cos{ \left( \frac{\pi}{2} \right) } \ln { (2-1) } - \frac{\sin{\frac{\pi}{2}}}{2-1}}{-\frac{\pi}{2}\sin{\frac{\pi}{2}}}} \\ &= e^{\frac{\frac{\pi}{2}\cdot 0 \cdot 0 - \frac{1}{1}}{-\frac{\pi}{2} \cdot 1}} \\ &= e^{\frac{2}{\pi}} \end{align*}
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  5. #5
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    Re: Finding Limit With L'Hospital's Rule

    Just take the derivative of the numerator and denominator if the first result is indeterminant.
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