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Homework Help: Finding Limit using Derivative

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Express the limit as a derivative and evaluate.

    limx-->pi/3 (Cos(theta) - 0.5)/(theta - pi/3)

    3. The attempt at a solution

    I derived the function using quotient rule and then didn't know what to do.

    (theta - pi/3)(-Sin(theta)) - (Cos(theta) - 0.5)

    (theta - pi/3)2
  2. jcsd
  3. Nov 3, 2009 #2


    Staff: Mentor

    You're missing the point of this exercise. You are supposed to interpret this limit as the derivative of some function, not take the derivative of it.

    Recall that the derivative can be defined as this limit:
    [tex]\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}[/tex]

    Hint: cosine of what = .5?
  4. Nov 3, 2009 #3
    Okay, thanks

    Does the wording of the question seem kind of weird to you? because I copied it word for word
  5. Nov 3, 2009 #4


    Staff: Mentor

    No, the wording seems fine to me.

    BTW, you don't "derive" a function, you differentiate it. I know it seems weird that you don't "derive" something to get its derivative, but that's the way it is.
  6. Nov 3, 2009 #5
    I got -[tex]\sqrt{3}[/tex]/2 as an answer, but I had to change the format of the derivative formula to this:

    limtheta-->pi/3 [tex]\stackrel{\underline{Cos(theta + h) - Cos(theta)}}{h}[/tex]

    But I don't know how to do it in the other format. Do you use conjugates, or what?
  7. Nov 3, 2009 #6


    Staff: Mentor

    Your answer is correct, so it seems you understand what the problem is asking for. Your revised limit needs some work, though, as it needs to be in terms of h going to 0, like so:
    [tex]\lim_{h \rightarrow 0} \frac{cos(\pi /3 + h) - cos(\pi /3)}{h}[/tex]
    which equals -sin([itex]\pi[/itex] /3).
  8. Nov 3, 2009 #7
    That's the correct answer, but I don't think that's what they're asking you to do, or at least how they want you to do the problem. Look at what Mark44 posted originally. That limit is equal to f'(a), which is f'(x) evaluated at a. Follow his hint to do it the original way. Either that or follow what he's posted above.
  9. Nov 3, 2009 #8
    yeah sry Mark that's what I meant.

    Anyways, I'm good on that one, but I'm still a little fuzzy on how to find this limit:

    limx-->(pi/3) [tex]\stackrel{\underline{(Cos(\theta) - 0.5)}}{(\theta - \pi/3)}[/tex]

    and I'm pretty sure that's what the question was asking.
    Last edited: Nov 3, 2009
  10. Nov 3, 2009 #9


    Staff: Mentor

    This is one version of the definition of d/d(theta)[cos(theta)], evaluated at theta = pi/3. Your limit should be as theta --> pi/3. No x.
  11. Nov 3, 2009 #10
    w/e about the x's and theta's that's just semantics.
    I know that it's a version of d/d(theta). I want to know how to find the limit using that.
  12. Nov 3, 2009 #11
    Well whether it's x or theta there changes the meaning of the problem completely. It's just a formula, but you have to sort of work backwards. An alternate definition of the derivative, which Mark44 already gave you, evaluated at the point a is
    [tex] f'(a) = \lim_{\theta\to a} \frac{f(\theta)-f(a)}{\theta-a}[/tex]
    You should be able to evaluate your limit from here.
  13. Nov 3, 2009 #12


    Staff: Mentor

    Doing well in mathematics and science and engineering is partly about paying attention to details ("semantics").
    Can you evaluate this limit?
    [tex]\lim_{h \rightarrow 0} \frac{cos(\pi /3 + h) - cos(\pi /3)}{h}[/tex]

    You can convert the other limit to this form by letting h = [itex]\theta - \pi /3[/itex]. Theta approaching pi/3 is equivalent to h approaching 0 in the limit above.
  14. Nov 4, 2009 #13
    Okay, thanks guys, sry if I sound frustrated

    Except I had already made it to that step just by reading the question. It gave it to me in this format:

    [tex] lim_{x->\pi /3} \stackrel{\underline{Cos(\theta ) - 0.5}}{\theta - \pi /3} [/tex]

    My problem isn't converting it to this, it's how to work with this formula to find the limit.
    It might seem obvious to you guys, but I'm missing some kind of technique because everytime I end up with 0/0 and I confuse myself.

    Help :confused:
  15. Nov 4, 2009 #14


    Staff: Mentor

    I think you are not realizing that cos(pi/3) = .5
  16. Nov 4, 2009 #15
    I am realizing that. That's why when I plug pi/3 into theta I get

    (0.5 - 0.5)/(pi/3 - pi/3)

    both are zero, so I get 0/0 and I freak out
  17. Nov 4, 2009 #16
    [tex]\begin{align*}\lim_{\theta\to\pi/3} \frac{\cos\theta-0.5}{\theta-\pi/3} &= \lim_{\theta\to\pi/3} \frac{\cos\theta-\cos(\pi/3)}{\theta-\pi/3} = (\cos\theta)' \,\big|_{\theta=\pi/3} \\
    &= -\sin\theta \,\big|_{\theta=\pi/3} = -\sin(\pi/3) \\
    &= -\frac{\sqrt{3}}{2}\end{align*}[/tex]
  18. Nov 4, 2009 #17
    Okay I get it.

    Thanks so much guys, sry
  19. Nov 4, 2009 #18
    No problem. What you were missing was using the definition of derivative that was posted to evaluate the limit. You kept trying to explicitly evaluate the limit instead of applying the definition in reverse.
  20. Nov 4, 2009 #19


    Staff: Mentor

    This is why you need limits, because you can't directly evaluate the quotient.
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