Continuity Applet Worksheet

To access the cross sections applet (based on work of Tom Banchoff) follow the link:
http://www.slu.edu/classes/maymk/banchoff/Continuity1Var.html

Spread the two graphs out where you can see them both, as well as the control panel.

We start with a polynomial function, which is obviously continuous.  This lets us look at what it means to say a graph is continuous at a point.  We then move on to more complicated functions.  The applet opens with the graph of the function f(x,y) = -x^4+x^2+x/4+1/2.  The applet has an f(x) graph window and a close up graph window.  Visually the graph of the function at x0 is a red point.  The graph of the function in a delta neighborhood of that point is given in purple.  Two blue lines are drawn epsilon above and below the point.  The definition of continuity says that no matter how small epsilon, delta can be chosen so that the purple curve is trapped by the blue lines.

1) The example starts with f(x) = -x^4+x^2+x/4+1/2, epsilon = 0.1, and delta = 0.5. Find a value of delta that works with this epsilon at this point.  (Make delta smaller until the purple curve is between between the two lines.)
Now reduce epsilon to 0.05 and find a delta that works for that epsilon and that point.
Change the value of x0 up and down to find the range of x values for which your delta works.
Repeat the process with epsilon = 0.01.

2) Next we work with a function that is clearly discontinuous.  Set f(x) = x-ipart(2*x)/2.  This gives a sawtooth function that jumps when x is some number of halves.
Set x0 = .6 and find values of delta  that work when epsilon is 0.5, 0.1, and 0.02.  In each case find a range of x values for which your delta works.
Set x0 = 1/2.  Explain why you can't find a value of delta that works if epsilon < .5.

3) Set of f(x) = x*sin(1/x) if x≠0, with f(0)=x.  Justify the claim that this function is continuous at x=0 by finding delta that work when epsilon is 0.5, 0.1, and 0.02.

4) Set of f(x) = sin(1/x) if x≠0, with f(0)=x.  Justify the claim that this function is not continuous at x=0 by finding a value of epsilon is for which no value of delta works.

Return to the Continuity Applet page.

Last updated By Mike May, S.J., January 25, 2006.