Problem of the Week #2

Problem Two: 1998-1999

Let the function f be defined by

(a) Prove that f is continuous at x = 0.
(b) Prove that (0) exists and find it by using a definition of the derivative.
(c) Prove that is not continuous at x = 0.

Solution: (a) Examine the graph of f to determine if it's continuous at x = 0:

(Figure One) Clearly, as far zoomed in as we are, this function is approaching a height of 0 as we approach x = 0. The sine curve is "damped" once I multiply by the x squared term. If the general limit exists as we approach 0, and it equals f(0), which it does, then f is dy definition continuous at x = 0.

(b) Originally, this problem said to use both definitions of the derivative. The difference quotient is quite difficult if we try to use it, so we use the alternative form, instead. Watch this:

(Figure Two) Therefore, the derivative is, in essence, this limit. How do we find the limit? Why not look at its graph?

(Figure Three) Clearly, the limit of this graph is 0 as x approaches 0. Because this limit represents the derivative of our original problem,

(0) = 0.

(c) This problem will be much easier once we learn more techniques for finding derivatives--namely the product rule and the chain rule. Giving a detailed solution now is unnecessary. For now, suffice it to say to follow the procedure we used in Figure Two a second time, using the result from Figure Two; the result will be sin (1/x), which oscillates infinitely as x approaches 0. Any function which acts in this fashion cannot have a general limit at that point.

Superbowl of Calculus