Copyright © Cengage Learning. All rights reserved. 3 Introduction to the Derivative

Copyright © Cengage Learning. All rights reserved. 3.2 Limits and Continuity

3 The graph in Figure 9 appears to have breaks, or discontinuities, at x = 0 and at x = 1. At x = 0 we saw that lim x→0 f (x) does not exist because the left- and right-hand limits are not the same. Thus, the discontinuity at x = 0 seems to be due to the fact that the limit does not exist there. Figure 9

4 Limits and Continuity On the other hand, at x = 1, lim x→1 f (x) does exist (it is equal to 1), but is not equal to f (1) = 2. Thus, we have identified two kinds of discontinuity: 1. Points where the limit of the function does not exist. 2. Points where the limit exists but does not equal the value of the function. x = 0 in Figure 9 because lim x→0 f (x) does not exist. x = 1 in Figure 9 because lim x→1 f (x) = 1 ≠ f (1)

5 Limits and Continuity On the other hand, there is no discontinuity at, say, x = –2, where we find that lim x→–2 f (x) exists and equals 2 and f (–2) is also equal to 2. In other words, lim x→–2 f (x) = 2 = f (–2). The point x = –2 is an example of a point where f is continuous. (Notice that you can draw the portion of the graph near x = –2 without lifting your pencil from the paper.) Similarly, f is continuous at every point other than x = 0 and x = 1. Here is the mathematical definition.

6 Limits and Continuity Continuous Function Let f be a function and let a be a number in the domain of f. Then f is continuous at a if a. lim x→a f (x) exists, and b. lim x→a f (x) = f (a). The function f is said to be continuous on its domain if it is continuous at each point in its domain.

7 Limits and Continuity If f is not continuous at a particular a in its domain, we say that f is discontinuous at a or that f has a discontinuity at a. Thus, a discontinuity can occur at x = a if either a. lim x→a f (x) does not exist, or b. lim x→a f (x) exists but is not equal to f (a). Quick Example The function shown in Figure 9 is continuous at x = –1 and x = 2. It is discontinuous at x = 0 and x = 1, and so is not continuous on its domain. Figure 9

8 Example 1 – Continuous and Discontinuous Functions Which of the following functions are continuous on their domains? x + 3 if x ≤ 1 x + 3 if x ≤ 1 5 – x if x > 1 1 – x if x > 1 if x ≠ 0 0 if x = 0 a. h(x) =b. k(x) = d. g(x) =c. f (x) =

9 Example 1 – Solution Solution for a and b. The graphs of h and k are shown in Figure 11. Figure 11

10 Example 1 – Solution Even though the graph of h is made up of two different line segments, it is continuous at every point of its domain, including x = 1 because = 4 On the other hand, x = 1 is also in the domain of k, but lim x→1 k(x) does not exist. Thus, k is discontinuous at x = 1 and thus not continuous on its domain. cont’d = h(1).

11 Example 1 – Solution Solution for c and d. The graphs of f and g are shown in Figure 12. cont’d Figure 12

12 Example 1 – Solution The domain of f consists of all real numbers except 0 and f is continuous at all such numbers. (Notice that 0 is not in the domain of f, so the question of continuity at 0 does not arise.) Thus, f is continuous on its domain. The function g, on the other hand, has its domain expanded to include 0, so we now need to check whether g is continuous at 0. From the graph, it is easy to see that g is discontinuous there because lim x→0 g(x) does not exist. Thus, g is not continuous on its domain because it is discontinuous at 0. cont’d