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. limit if and only if for every epsilon greater than 0, there exists, a delta greater than 0, such that, 0 less 
than |x - c|less than delta guarantees that |f(x) - L|less than epsilon.

So what is this saying? Let's start with 0 less
than |x - c|less than delta. The absolute value is around x - c because we want x to be within delta 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|<delta if and only if(iff)
-(x-c)<delta and (x-c)<delta iff
x>c-delta and xless than c+delta iff
c-delta less than x less than c+delta.

What about the |f(x) - L|less than epsilon ? Using the above reasoning and solving for f(x), we can say f(x) lies within epsilon 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 |f(x) - L|. We specify how much this distance changes, by choosing a delta which also affects the epsilon. As delta gets larger, the distance |x - c| and epsilon are getting larger which is making |f(x) - L| proportionately larger. As delta gets smaller, the distance |x - c| is getting smaller and smaller, which makes |f(x) - L| and epsilon proportionately smaller. If for any delta we can find a corresponding epsilon, then we say limit f(x).


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:

limit of 
f(x)

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

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

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

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


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

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

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


limit of f(x).