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Math Help - l'hopital Question

  1. #1
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    l'hopital Question

    Hiii :3

    I'm stuck on this question:


    lim (x->1) (2-x)^ tan(π/2)x


    I did the ln thing but now i'm stuck what do I do next?
    Plug it in?



    Thank youuus :]
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  2. #2
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    Quote Originally Posted by TeriyakiDonnQ View Post
    Hiii :3

    I'm stuck on this question:


    lim (x->1) (2-x)^ tan(π/2)x


    I did the ln thing but now i'm stuck what do I do next?
    Plug it in?



    Thank youuus :]
    Seeing as you've been told to use L'Hospital's Rule, it would make sense to use L'Hospital's Rule, wouldn't it?


    You are correct that you should apply the exponential-logarithmic transformation...

    \lim_{x \to 1} (2 - x)^{\tan{\frac{\pi x}{2}}} = \lim_{x \to 1}e^{\ln{(2 - x)^{\tan{\frac{\pi x}{2}}}}}

     = \lim_{x \to 1}e^{\tan{\left(\frac{\pi x}{2}\right)}\ln{(2 - x)}}

     = \lim_{x \to 1}e^{\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{(2 - x)}}{\cos{\left(\frac{\pi x}{2}\right)}}}

     = e^{\lim_{x \to 1}\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{(2 -  x)}}{\cos{\left(\frac{\pi x}{2}\right)}}}.


    This limit is now of the form \frac{0}{0}, so now you can use L'Hospital's Rule.
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