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Finding the limit of a function

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose y = 9(x - 3) + 30 is the tangent line to f(x) at x = 3.

    The limit I need to find:

    http://img259.imageshack.us/img259/2963/b3275332191bc2fc8561357mt3.gif

    2. No Relevant equations

    3. The attempt at a solution

    Well I know that f(3) is equal to 30. So that means that the numerator will be 30 - 30 and the denominator will be 0. So I can't remember what I need to do to find the exact answer. Is there some way that I can determine what f(x) is? I know f'(3) = 9 and f(3) = 30.

    I could work backwards and say f(2) is 21, f(1) is 12, f(0) is 3 but that would require that f be linear.

    How can I find this limit and be sure that it is the right answer?
     
  2. jcsd
  3. Sep 30, 2008 #2
    You are asked the derative in x=3 but you know the tangent line in x=3 so we know the derative, right?
     
  4. Sep 30, 2008 #3

    HallsofIvy

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    One way of defining "derivative" of f(x) at x= a is that the derivative is the slope of the tangent line to the graph of y= f(x) at x= a. The question is much simpler than you think it is: no calculation required.
     
  5. Sep 30, 2008 #4
    Well I know the slope of the tangent line at 3 is 9. But the limit at 3 is 30.

    I think what I don't understand is how to evaluate the limit. (Normally evaluating the limit is the same as plugging in a value into the function, but when it is like this and everything is divided by 0 I get confused.) I don't understand how the derivative at a point affects the limit of an expression.

    Maybe the 0 in the denominator is what is causing me all this trouble and really the limit is +/- infinity?
     
  6. Sep 30, 2008 #5

    HallsofIvy

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    You can't take the limit because you don't know what f is!

    So what limit are you talking about being 30? Yes, the function y= 9(x-3)+ 30 has value 30 at x= 3. That tells you that, in order that that line be tangent to the graph of y= f(x), f(3) must be 30. That has nothing to do with the fact that f'(3)= 9 and surely you know that an equivalent way of defining the derivative (equivalent to 'slope of the tangent line') is
    [tex]f'(3)= \lim_{h\rightarrow 0}\frac{f(3+h)- f(3)}{h}[/itex]

    You know, from the tangent line given, that f'(3)= 9. You know from the "limit definition" of the derivative that the limit given is the derivative of f at x= 3. Therefore, that limit is 9.

    The fact that the denominator becomes 0 does NOT tell you that a limit must be [itex]\pm\infty[/itex]; not if the numerator goes to 0 also- which it always does in a derivative limit: the numerator is f(a+ h)- f(a) which becomes f(a)- f(a)= 0. Again, you cannot directly calculate that limit because you don't know what f(x) is.
     
  7. Sep 30, 2008 #6
    Okay, so this makes sense now. Really the key to the problem was the limit definition of derivative.
     
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