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Limit-derivative question

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Given y = sin x + cos x , evaluate limit h approaches 0 of [(f(pi+h)-f(pi)]/h


    2. Relevant equations
    d/dx (sin x) = cos x
    d/dx (cos x) = -sin x

    3. The attempt at a solution
    This was a question on my quiz today. Its answer, which I knew from the glance, had to be the derivative of the equation at x = pi.
    Using d/dx rule,
    dy/dx = cos x - sin x
    at x = pi; cos pi - sin pi = -1 - 0 = -1
    I found the answer by using derivative rules but I couldn't do it using the limit concept.

    lim h->0 [(sin pi+h)+(cos pi+h)-(sin pi)-(cos pi)]/h
    lim h->0 [(sin pi)(cos h)+(cos pi)(sin h)+(cos pi)(cos h)-(sin pi)(sin h)-(sin pi)-(cos pi)]/h
    lim h->0 [0-(sin h)-(cos h)-0-0+1]/h
    lim h->0 [1-(sin h)-(cos h)]/h

    That's as far as I've got. I couldn't cancel the h in the denominator. Please help, thank you.
     
  2. jcsd
  3. Oct 26, 2009 #2

    lurflurf

    User Avatar
    Homework Helper

    [1-(sin h)-(cos h)]/h=[1-cos h]/h-sin h/h
    so consider
    lim_h-> [1-cos h]/h=cos'(0)=0
    lim_h->0 sin h/h=sin'(0)=1
    So to avoid using derivatives you will need to find those limits another way in fact trigonometric identities can be used to avoid lim_h-> [1-cos h]/h which leaves
    lim_h->0 sin h/h=sin'(0)=1
     
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