# Limits Part II

In the previous section we discussed the general idea of a limit, and evaluated some limits of a graphed function. However, looking at the graph of a function is not any way to formally work with limits. In this section, we are going to define a limit and work an epsilon-delta proof that shows a limit for a specific function exists at a particular point.

Suppose we have a function and it exists and is defined around a particular point, c. if and only if for every 0, there exists, a 0, such that, 0 |x - c| guarantees that | - L| .

So what is this saying? Let's start with 0 |x - c| . The absolute value is around x - c because we want x to be within units of c, both to the left and to the right of c. You can see what we mean by this by solving for x:

|x-c|< if and only if(iff)
-(x-c)< and (x-c)< iff
x>c- and x c+ iff
c- x c+.

What about the | - L| ? Using the above reasoning and solving for , we can say lies within units of L, both to the left and to the right of L.

The two statements are connected together by x which is approaching(but not necessarily equal to), c. As the distance |x - c| changes this affects the value of | - L|. We specify how much this distance changes, by choosing a which also affects the . As gets larger, the distance |x - c| and are getting larger which is making | - L| proportionately larger. As gets smaller, the distance |x - c| is getting smaller and smaller, which makes | - L| and proportionately smaller. If for any we can find a corresponding , then we say .

Now we know for a limit to exist we can find an epsilon for every delta. Let's try to do an epsilon-delta proof on a real limit:

Since this is true, the following statement is also true which comes from the definition of a limit:

if 0 < |x - 5| < then |(5x + 10) - 35| < .

All we did was substitute the numbers in the example for x, c, , and L into the definition of a limit.

Right now it's not very clear what value of corresponds to . Let's try to rework the numbers in the inequality involving to make it look more like the inequality involving .

|(5x + 10) - 35| < iff |5x - 25| < iff |5(x - 5)| < iff
5|x - 5| < iff |x - 5| < /5.

Now we have the inequality involving looking exactly like the inequality involving . It should be clear that = /5.

What we have done is find an for every . If = 2, then = 2/5. If = 25, then = 5. If = 100,000, then = 20,000. Since we have found an for every , we can safely say

.