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Homework Help: Limit as h->0 when [cos(pi/2+h)]/h

  1. Oct 22, 2004 #1
    limit as h->0 when [cos(pi/2+h)]/h

    when I sub in zero...I get 0/0...so I have to use L'hopital's rule
    So I take the derivative of the numberator and denominator...

    =[-sin(pi/2+h)*(0+1)]/1
    =-sin(pi/2+h)
    =-sin(pi/2+0)
    = -1

    is this correct?
     
  2. jcsd
  3. Oct 22, 2004 #2
    First off do you mean

    [tex]\cos(\frac{\pi}{2} + h)[/tex]
    or
    [tex]\cos(\frac{\pi + h}{2})[/tex]

    I think you mean the first expression

    Note that [tex]\cos(\frac{\pi}{2} + h) = -\sin(h)[/tex]
    and the given limit equals
    [tex]-\lim_{h->0} \frac{\sin h}{h} = -1[/tex]

    At this juncture you may use L Hospital's Theorem to reduce it to cos(h)/h but the fraction sin(h)/h goes to 1 in the limit as can be proved without L-Hospital's Theorem.

    Your solution is correct.

    Cheers
    Vivek
     
    Last edited: Oct 22, 2004
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