Intervals of Continuity Even though not all functions are continuous on all real numbers (i.e., everywhere), we can still talk about the continuity of a function in terms of intervals. As you may recall, there are three kinds of intervals: open intervals, closed intervals, half-open (or half-closed) intervals. Definition of f(x) is continuous on an open interval: We say a function f(x) is continuous on an open interval of (a, b) if there are no points of discontinuity on the interval (a, b). That is, f(x) is continuous at any x where a < x < b. We say a function f(x) is continuous on a closed interval of [a, b] if i. f(x) is continuous on the open interval (a, b), ii. f(a) and f(b) are defined, and iii. limxa f(x) = f(a) and limxb f(x) = f(b). We say a function f(x) is continuous on a half-open interval of [a, b) if i. f(x) is continuous on the open interval (a, b), and ii. f(a) is defined and limxa f(x) = f(a). We say a function f(x) is continuous on a half-open interval of (a, b] if i. f(x) is continuous on the open interval (a, b), and ii. ______________________________. b a a b b a b a b a + – b a b a +

Continuous on its Domain Recall that functions such as f(x) = x1/2 and f(x) = log x are not continuous on all real numbers, nevertheless, they are continuous at every number in their domains. For example, the domain of f(x) = x1/2 is [0, ) and its interval of continuity is also [0, ). Similarly, the domain of f(x) = log x is (0, ) and its interval of continuity is also (0, ). Another example is f(x) = 1/x, where its graph has two pieces and there is a vertical asymptote at x = 0. However, since its domain is all real numbers except 0, i.e., (, 0)  (0, ), and we can see that the left piece of f(x) = 1/x is continuous on (, 0) where as the right piece is continuous on (0, ), we still say that the function is continuous on its domain. So what is an example of a function which is not continuous on its domain? There are many functions which are not continuous on their domains. For example, the integer function, f(x) = [x]. The domain is all real numbers (plug any real number x into the integer function, it will yield back the greatest integer ≤ x). However, the function is not continuous on its domain since it is discontinuous at every integer. Another example is the sign function, f(x) = sgn x = where given any real number x, this function will yield back 1 if x is positive, 1 if x is negative, and 0 is x is zero. As you can see, its domain is all real numbers, but it’s not continuous at x = 0.

How to Give the Intervals of Continuity of a Function If a function is continuous on all real numbers, e.g., f(x) = x2, then we say its interval of continuity (IOC) is (, ). If a function is continuous on its domain, then its interval(s) of continuity is same as its domain. For example, i) f(x) = x1/2  IOC = [0, ); ii) f(x) = 1/x  IOC = (, 0)  (0, ). If a function is not continuous on its domain, e.g., f(x) = sgn x, we must say the IOC is (, 0)  (0, ) despite that the domain is (, ). This implies the interval notation for the domain is not necessarily same as the intervals of continuity. Example For the function with graph below, give the domain and intervals of continuity using the least number of intervals as possible. Domain: ____________________________ IOC: _______________________________ 3 y = f(x) 2 1 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 9 10 11 –1 –2 –3 Domain: (-oo, -1) U (-1,5) U (5, oo) IOC: (-oo, -4] U (-4,-1) U (-1, 1)U (1, 3) U (3, 5) U (5, 8) U [8.oo)

Properties of Functions which are Continuous on their Domains Theorem: The following types of functions are continuous at every number in their domains: i) polynomial functions ii) rational functions iii) root functions iv) trigonometric functions v) exponential functions vi) logarithmic functions vii) absolute-value functions viii) inverse trigonometric functions Properties: If f(x) and g(x) are two functions continuous at every number in their respective domains, and let D be the domain of f(x) + g(x) and E be the domain of f(g(x)), then: i)f(x) + g(x), f(x) – g(x) and f(x)∙g(x) are continuous on D, ii)f(x)/g(x) is continuous on D – {x | g(x) = 0}, and iii)f(g(x)) is continuous on E. Continuous on all real numbers Continuous only on their domains Polynomial Rational Odd-indexed root Even-indexed root sin, cos tan, cot, sec, csc tan–1, cot–1 sin–1, cos–1, sec–1, csc–1 Exponential Logarithmic Absolute-value Example Let f(x) = sin x and g(x) = x2 – 1 . Find the intervals of continuity of the following functions: 1. f(x) + g(x) 2. f(x) – g(x) 3. f(x)∙g(x) 4. f(g(x)) 5. g(f(x)) 6. f(x)/g(x) 7. g(x)/f(x)

How Do We Find the Limit of a Function Without the Graph? There are several ways we can find the limit of a function f as x approaches a without knowing its graph. One obvious way is to use numbers close to a. See the following examples: 1. limx1 (x2 + 1) = ___ 2. limx2 1/x2 = ___ x 1.9 1.99 1.999  2  2.001 2.01 2.1 1/x2 x 0.9 0.99 0.999  1  1.001 1.01 1.1 x2 + 1 3. limx3 (x – 3)/|x – 3| = ___ 4. limx2 1/(x2 – 4) = ___ x 2.9 2.99 2.999  3  3.001 3.01 3.1 x 1.9 1.99 1.999  2  2.001 2.01 2.1 5. limx–1 5/(x + 1) = ___ 6. limx0 6 sin x/x = ___ (where x in radians) x –1.1 –1.01 –1.001  –1  –.999 –.99 –.9 x –.1 –.01 –.001   0.001 0.01 0.1 7. limx2– 7/(x – 2) = ___ 8. limx2+ 8/(x + 2) = ___ 9. limx∞ 3x/(2x – 9) = ___ x 1.9 1.99 1.999  2 x 2  2.001 2.01 2.1 x 100 1,000 10,000  ∞

Implication of Continuity on Limits Recall that if f(x) is continuous at x = a, then the limit of f(x) as x approach a must exist and it must equal to f(a). Therefore, when we need to evaluate limxa f(x) and if we know f(x) is continuous at x = a, all we need to do is to evaluate f(a), i.e., whatever f(a) is, is the limit! Direct Substitution Property: If f(x) is continuous at x = a, then limxa f(x) = f(a). Rule of Thumb of Evaluating Limit When you evaluate the limit of a function f(x) as x approaches a, just plug in a into the f(x) first. That is, when limxa f(x) is asked, just do f(a). With this property, it allows us to evaluate the limit by using the so-called “plug-in” method. That is, as long as we know f(x) is continuous at a, even if we don’t know the graph of f(x), we can just evaluate f(a) and that will be limit! We don’t need to use any numbers close to a, hence, it is a much better and faster way of finding the limit than the “tabular” method used on page 5. Examples: 1. limx1 (x2 + 1) = 2 2. limx2 1/x2 = 1/22 = ¼ 3. limx3+ (x2 – 2)/(x + 1) = 7/4 4. limx– cos x = The rule of thumb above really is another way (actually, my way) to state the Direct Substitution Property. The difference is: here we don’t even need to care whether f(x) is continuous at a or not. By Direction Substitution Property By My Rule of Thumb Example 1: x2 + 1 is a quadratic function. Quadratic functions are a type of polynomial functions and polynomial functions are continuous everywhere, therefore it’s okay to plug the 1 into x, and obtain 12 + 1 = 2 as the limit. Example 1: 12 + 1 = 2 Example 2: 1/x2 is a rational function with a vertical asymptote at x = 0. So 1/x2 is discontinuous at x = 0 but it’s continuous elsewhere, therefore it’s okay to plug the 2 into x, and obtain 1/22 = ¼ as the limit. Example 2: 1/22 = 4