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]<br /> <br /> You know, from the tangent line given, that f'(3)= 9. You know from the "limit definition" of the derivative that the limit given <b>is</b> the derivative of f at x= 3. Therefore, that limit is 9.<br /> <br /> 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 <b>cannot</b> directly calculate that limit because you don't know what f(x) is.[/tex]

 

[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]<br /> <br /> You know, from the tangent line given, that f'(3)= 9. You know from the "limit definition" of the derivative that the limit given <b>is</b> the derivative of f at x= 3. Therefore, that limit is 9.<br /> <br /> 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 <b>cannot</b> directly calculate that limit because you don't know what f(x) is.[/tex]

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