Finding the limit of a function

[altegron]

Homework Statement

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

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?

 

[dirk_mec1]

You are asked the derative in x=3 but you know the tangent line in x=3 so we know the derative, right?

 

[HallsofIvy]

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.

 

[altegron]

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?

 

[HallsofIvy]

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 $\pm\infty$; 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.

 

[altegron]

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 $\pm\infty$; 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.

Okay, so this makes sense now. Really the key to the problem was the limit definition of derivative.