# Finding the limit of a function

1. Sep 30, 2008

### altegron

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 [Broken]

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?

Last edited by a moderator: May 3, 2017
2. Sep 30, 2008

### 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?

3. Sep 30, 2008

### HallsofIvy

Staff Emeritus
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.

4. Sep 30, 2008

### 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?

5. Sep 30, 2008

### HallsofIvy

Staff Emeritus
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.

6. Sep 30, 2008

### altegron

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