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Evaluating a limit to 2 decimal places

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data

    If f(x) = sin x, evaluate

    lim f(2+h) - f(2) / h
    h->0

    Evaluate to 2 decimal places


    2. Relevant equations



    3. The attempt at a solution

    I think that since f(2) then the answer is sin 2 which is .91

    What do you guys think?
     
  2. jcsd
  3. Jan 15, 2009 #2

    danago

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    If it said lim sin(h) as h->2, then the answer would be sin 2 since sin(x) is continuous.

    However, the limit expression is not sin(x), it is f(2+h) - f(2) / h as h->0. Does this limit look familiar? What does it define?
     
  4. Jan 15, 2009 #3

    HallsofIvy

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    But I suspect that this problem was intended to lead to the derivative of sin x and so that method would be inappropriate. What if you just try small values of h in
    [tex]\frac{sin(2+h)- sin(2)}{h}[/tex] and see how small h has to be to so that you get the same answer to 2 decimal places?
     
  5. Jan 15, 2009 #4
    Halls, I actually did the equation like you have it there, and then using the sum rule for sine, sinacos b - cos asinb I got a final answer of cos 2 which equalled -0.42. I havent plugged in any values to check my work, i just fiddled with the equation.
     
  6. Jan 15, 2009 #5

    HallsofIvy

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    All the problem asked is that you plug in values!

    What I meant was, since sin(2)= 0.9093,
    Taking h= 0.1, (f(2+h)- f(2))/h= (sin(2.1)- sin(2))/.1= -.4609.
    Taking h= 0.01, (f(2+h)- f(2))/h= (sin(2.01)- sin(2))/.01= -.4206
    Taking h= 0.001, (f(2+h)- f(2))/h= (sin(2.001)- sin(2))/.001= -.4166
    Taking h= 0.0001, (f(2+h)- f(2))/h= (sin(2.0001)- sin(2))/.0001= -.4162
     
  7. Jan 15, 2009 #6
    I did it the way I did it because there was an example in my manual that led me to do it that way. I did it your way too and both ways led me to the same answer.

    Thanks for your help!
     
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