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Limit problem

  1. Dec 8, 2012 #1
    Hello.
    I have been trying to find this limit:

    Lim as x --> 1 of (sin((1-x)/2)*tan(Pi*x/2))

    Of course I dont want to solve it using L'Hospital. I have tried several ways but ended up in one of these.

    lim as y -->0 of (y*tan(pi/2-y*pi))

    The answer when using L'Hospital is 1/pi.

    How can I go about getting the same answer without using L'Hospital. Thanks.
     
  2. jcsd
  3. Dec 8, 2012 #2

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    = ##\sin\left(\frac{1-x}{2}\right) \frac{\sin(\pi x/2)}{\cos(\pi x/2)}##
    sin(πx/2) goes to 1 and does not matter. With z=x/2-1/2 and z->0 this can be rewritten as
    $$ \frac{-\sin\left(z\right)}{\cos(\pi z+\frac{\pi}{2})} = \frac{\sin\left(z\right)}{\sin(\pi z)}$$
    You can use common upper and lower bounds for sin() to get the limit of that expression.
     
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