Math 1215 with Maple --- Lab 2

Math 1215 with Maple: Wednesday, September 23, 1998

The epsilon-delta Definition of Limit; Continuity

Assignment: Exercises 2 - 4, Exercises 6 - 8.

Examples

(From Varberg & Purcell Technology Project for Chapter 2)

Example 1: Graph the piecewise-defined function

         f(x) =  x^3 + 2       if x <= -1
                
                 x^4 + 1       if x >=  1

                 x^2 - x + 1   elsewhere

For which values of x does f fail to be continuous?

Solution: There are several ways to graph such a function in Maple, say on the interval from -2 to 2. One is shown below. Note: The first four of these commands ends with a colon rather than a semicolon to to avoid printing long lists of points to be plotted on your screen. The final command using display ends with a colon because we want to see the plot on our screen.

[> with(plots):        #Load the plots package
[> p1 := plot( x^3 + 2, x = -2..-1):
[> p2 := plot( x^4 + 1, x = 1..2):
[> p3 := plot( x^2 - x + 1, x = -1..1):
[> display({p1, p2, p3});

We see that the graph of f comes in 3 pieces, with a jump at x = -1 and at x = 1. Thus

  lim    f(x) and     lim  f(x)
x -> -1             x -> 1

will not exist, because the left and right limits at these points do not match. So f is not continuous at these points. Suppose you want to redefine f to make it continuous by substituting a different parabolic patch in the middle. In other words find numbers b and c so that

         f(x) =  x^3 + 2       if x <= -1
                
                 x^4 + 1       if x >=  1

                 x^2 + bx + c   elsewhere

will be continuous. For continuity we need the endpoints of the middle piece to match up with the endpoints of the other two pieces so there will be no jumps in the graph. Since

f(-1) = (-1)^3 + 2 = 1 we need  (-1)^2 + b(-1) + c = 1

and since

f(1) = (1)^4 + 1 = 2 we must also have  1^2 + b(1) + c = 2

So we can use Maple to find b and c:

[> solve( {1 - b + c =1, 1 + b + c = 2});

Example 2: We can also use Maple to graph rational functions and factor polynomials. For example to find

  lim          x^3 - 9x^2 - 45x - 91
x -> 13       ----------------------
                      x - 13

we can factor the numerator, or simplify the rational expression and substitute x = 13, or graph the rational function near x = 13:

[> p := x -> (x^3 - 9*x^2 - 45*x - 91)/(x-13);
[> p(13);                                       #plugging in fails
[> numer(p(x));    factor( numer(p(x)) );
[> s := simplify( p(x) );
[> subs( x = 13, s );
[> plot( p(x), x = 12..14 );

Intuitively we see that

  lim          x^3 - 9x^2 - 45x - 91
x -> 13       ----------------------   =  228
                      x - 13

If this is true then given any epsilon > 0 we can find a corresponding delta > 0 so that

 
   | x^3 - 9x^2 - 45x - 91       |
   |---------------------- - 228 |   <  epsilon
   |        x - 13               |

whenever  0 < | x - 13 | < delta.

Graphically, we want to see how to choose x in a small interval centered at x = 13 so that all the corresponding y values will lie in the interval (228 - epsilon, 228+epsilon). Suppose that epsilon = 1.0. Then we can plot to find the corresponding delta:

[> epsilon := 1.0;
[> plot({ p(x), 228 + epsilon,
           228 - epsilon}, x = 12..14 );

It looks like delta is about 0.1 and so we might replot on a smaller interval to see this better:

[> epsilon := 1.0;
[> plot({ p(x), 228 + epsilon,
           228 - epsilon}, x = 12.8..13.2 );

If we click on the points where these graphs cross we find that the intersections are at (12.97, 227) and (13.03, 229) so delta = anything smaller that 0.03 should work for epsilon = 1.

Exercises

Work through Example 1 above. Find values for b and c and graph the redefined continuous function.
Work through Example 2 above. When you print out your graphs, draw in by hand and label the vertical lines that correspond to delta for epsilon = 1.
(Also from Varberg and Purcell, Technology Project, end of Chapter 2)
Investigate
```
    lim     sqrt( 25 + 3x ) - sqrt( 25 - x )
           ---------------------------------
   x -> 0                x
```
by graphing this function near x = 0. Based on your graph, does this limit exist? What is its value? Find delta that corresponds to epsilon = 0.01. Repeat for epsilon = 0.001.
Compute this limit algebraically by rationalizing the numerator. (Do this by hand if you prefer.)
Consider the function f(x) = sin(x) / x. We can investigate
```
  lim       sin(x)
            ------
x -> 0        x
```
by plotting on intervals that are closing in on x = 0. Plot this function for x between -1 and 1. Based on this graph, does the limit exist? What is its value? Now using the graphical method discussed above, find a delta that corresponds to epsilon = 0.1. Repeat for epsilon = 0.01.
Investigate
```
  lim       sin(mx)
            -------
x -> 0         x
```
for m = 1, 2, 3, 4, 5 ... by graphing for x from -1 to 1.
Graph the function f(x) = sin(1/x) for x between -1 and 1. Now replot for x between -0.1 and 0.1. Replot again, now for x between -0.01 and 0.01. In each case what is the maximum y-value in your plot? What is the minimum? Discuss this sequence of plots and what it suggests about
```
   lim       sin(1/x)
  x -> 0
```
Use graphical methods to investigate
```
 lim        x sin(1/x)
x -> 0
```
If possible, find delta that corresponds to epsilon = 0.1 and to epsilon = 0.01. Provide graphical evidence to support your conclusions.
Given that
```
  lim       (1 + x)^(1/x)
 x -> 0
```
exist, estimate it accurately to 3 decimal places. You can do this numerically or graphically, by plotting on smaller and smaller intervals centered at x = 0. Do you recognize this important number?