Finding the largest delta

that works with a given epsilon

Example 1

Given . As x gets closer to 2, f(x) gets closer to 6. Find so that if then

Solution .

This means that we must find a such that if x is within units of 2, then f(x) is within .5 units of 6. In other words if x is between and then f(x) is between 5.5 and 6.5.

We will first try to find by trial and error. Below is a graph with

In this graph the lines and are drawn in green while the lines y = 6 - .5 = 5.5 and y = 6 + .5 = 6.6 are drawn in red. From this graph we can see that there are values of x within units of a = 2 (between the two green lines) that do not have corresponding y values that are within 6 - .5 and 6 + .5 (between the two blue lines).

Now let's try

Again there are x values which are close to a = 2 whose y values are outside of the red lines. Let's zoom in.

From this view we can see that we must have a smaller value for Next we try

There are still values of x within the green lines whose corresponding y values are more than .5 units away from L = 6.

Now we try (We will show how we got this value in the next example.)

We have success. Now every x value within the green lines has its corresponding y value within .5 units of 6. (i.e within the red lines).

Example 2

Using the function in example one, find the largest value of so that if then

Solution .

Graph y = f(x), y = 6+.5 = 6.5 and y = 6 - .5 = 5.5.

Because each horizontal line crosses the parabola twice, the equations, f(x) = 5.5 and f(x) = 6.5, will each have two solutions. Solving f(x) = 5.5 we get

Because we are taking the limit as x approaches 2, we ignore the negative root and let

Now solving f(x) = 6.5 we get

Since we want the result near x = 2, let

If we take x between and f(x) will between 5.5 and 6.5

To find delta we must find the distance from 2 to and the distance from 2 to

Let = 2 - 1.935962184 = .064037816

and = 2.061100044 - 2 = .061100044

Since these values are not the same, 2 is not in the center of the interval from to To get delta take the minimum of and which in this case is .061100044.

The following graph shows that each x that is within .061100044 of two has its corresponding y value within .5 of 6.

If we had used .064037816 as delta, then 2.064 would be within .0604037816 of two but

f(2.064) = 6.524288 would be more than .5 units from 6.

Example 3

Find the largest delta so that if then

Solution .

We need to find delta so that if x is between and then is between 8.01 and 7.99. Below is a zoomed view of the graphs of , y = 8.01 and y = 7.99

Solving

We get

Solving

We get

Let = .000833681 and = .000832986

The largest delta that we can use with epsilon of 0.01 is the minimum of and

This is confirmed in the graph below.