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Homework Help: Limit Question

  1. Dec 14, 2011 #1
    lim x → 0 sin(x)/x = 1

    This doesnt mean that f '(0) = 1 if f(x) = sin(x)/x does it?
  2. jcsd
  3. Dec 14, 2011 #2


    Staff: Mentor

    No, it's just a limit that's not related to a derivative.
  4. Dec 14, 2011 #3
    But derivates and limits are similar correct?
  5. Dec 14, 2011 #4


    Staff: Mentor

    A derivative is a limit, but a limit is not necessarily a derivative.
  6. Dec 14, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper

    That's a difference quotient for f(x)=sin(x). Not for f(x)=sin(x)/x.
  7. Dec 15, 2011 #6
    Definition of a derivative:

    df/dx = limΔx→0 [f(x+Δx)-f(x)]/Δx

    Changing the variables around, we find:

    df/dy = limx→0 [f(y+x)-f(y)]/x

    If limx→0 f(y+x)-f(y) = sin(x), then your relation holds. Not really sure what function would satisfy this though. Maybe you can tell me. :P
  8. Dec 15, 2011 #7


    User Avatar
    Science Advisor

    It certainly can be used to show that:

    [tex]\frac{dsin}{dx}(0)= \lim_{x\to 0}\frac{sin(x)- sin(0)}{x}= \lim_{x\to 0} \frac{sin(x)}{x}[/tex].
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