MATH 12002 - CALCULUS I §1.5: Continuity

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 12 Definition of Continuity

Intuitively, a function is continuous at a point if the graph can be drawn through the point without lifting your pencil; that is, there are no holes or jumps in the graph at that point.

This is not precise enough to accurately define continuity, however.

We will use the following definition: Definition A function y = f (x) is continuous at x = a if f (a) is defined and

lim f (x) = f (a). x→a

A consequence of this definition is that when x is near a, a “small” change in x results in a “small” change in y, and therefore no break or jump in the graph is possible.

D.L. White (Kent State University) 2 / 12 Definition of Continuity

We will use the following checklist to determine if a given function y = f (x) is continuous at a given point x = a:

1 Is f (a) defined? 2 Does lim f (x) exist? x→a 3 Does lim f (x) = f (a)? x→a

If the answer to each of these is YES, then f is continuous at x = a. If the answer to any one of these is NO, then f is not continuous at x = a.

D.L. White (Kent State University) 3 / 12 Examples

Example Determine whether the function x2 − 4 f (x) = x − 2 is continuous at x = 2 and justify your answer.

Solution Notice that when x = 2, the denominator of the function is 0. Therefore f (2) is not defined, and so f is NOT continuous at x = 2.

(x−2)(x+2) Note: Even though f (x) = x−2 , we cannot “cancel” the x − 2 factors in this situation. The function f (x) is undefined at x = 2 (and has a hole in its graph at x = 2), but the function g(x) = x + 2 has g(2) = 4. Cancelling changes the function; f (x) 6= g(x).

D.L. White (Kent State University) 4 / 12 Examples

Example Determine whether the function

 2  x − 4  if x 6= 2 f (x) = x − 2   3 if x = 2

is continuous at x = 2 and justify your answer.

D.L. White (Kent State University) 5 / 12 Examples

Solution 1 By the definition of the function, f (2) = 3, so f is defined at x = 2. 2 We have x2 − 4 lim f (x) = lim , since x 6= 2, x→2 x→2 x − 2 (x − 2)(x + 2) = lim x→2 x − 2 = lim (x + 2), since x 6= 2, x→2 = 2 + 2 = 4,

and so lim f (x) exists. x→2 3 Since f (2) = 3 6= 4 = lim f (x), we have lim f (x) 6= f (2), and so f is x→2 x→2 NOT continuous at x = 2.

D.L. White (Kent State University) 6 / 12 Examples

Example Determine if  x2 + 3x + 1 if x < 1 f (x) = 4 2 2x + 6x − 3 if x > 1 is continuous at x = 1 and justify your answer.

D.L. White (Kent State University) 7 / 12 Examples

Solution 1 We have f (1) = 2(14) + 6(12) − 3 = 2 + 6 − 3 = 5, and so f (1) is defined. 2 To determine if lim f (x) exists, we compute the one-sided limits. x→1

lim f (x) = lim (x2 + 3x + 1) = 12 + 3 · 1 + 1 = 5 x→1− x→1− and

lim f (x) = lim (2x4 + 6x2 − 3) = 2 · 14 + 6 · 12 − 3 = 5. x→1+ x→1+ Since lim f (x) = lim f (x), lim f (x) does exist. x→1− x→1+ x→1 3 Since lim f (x) = lim f (x) = 5, we have lim f (x) = 5 = f (1), and x→1− x→1+ x→1 so f is continuous at x = 1.

D.L. White (Kent State University) 8 / 12 Examples

Example Determine if  2  x + 3x + 1 if x < 1 g(x) = 7 if x = 1  2x4 + 6x2 − 3 if x > 1 is continuous at x = 1 and justify your answer.

D.L. White (Kent State University) 9 / 12 Examples

Solution 1 By the definition of the function, we have g(1) = 7, and so g(1) is defined. 2 To determine if lim g(x) exists, we compute the one-sided limits. x→1

lim g(x) = lim (x2 + 3x + 1) = 12 + 3 · 1 + 1 = 5 x→1− x→1− and

lim g(x) = lim (2x4 + 6x2 − 3) = 2 · 14 + 6 · 12 − 3 = 5. x→1+ x→1+ Since lim g(x) = lim g(x), lim g(x) does exist. x→1− x→1+ x→1 3 Since lim g(x) = lim g(x) = 5, we have lim g(x) = 5 6= 7 = g(1), x→1− x→1+ x→1 and so g is NOT continuous at x = 1.

D.L. White (Kent State University) 10 / 12 Polynomials and Rational Functions

Continuity for polynomials and rational functions is very easy to determine. From §1.4, we have Direct Substitution Property If f is a polynomial or rational function and a is in the domain of f , then

lim f (x) = f (a). x→a

Notice that this condition is precisely the definition of continuity at the point x = a, and so we have Theorem If f is a polynomial or rational function and a is in the domain of f , then f is continuous at x = a. This means that if f is a polynomial, then f is continuous at x = a for every real number a, and if f is a rational function, then f is continuous at x = a if and only if the denominator of f is not 0. D.L. White (Kent State University) 11 / 12 Polynomials and Rational Functions

Example Determine the values of x at which the function x2 + x − 6 f (x) = x2 + 2x − 3 is not continuous.

Solution Since f is a rational function, it is discontinuous precisely at the values of x for which the denominator is 0. The denominator is

x2 + 2x − 3 = (x + 3)(x − 1),

and so is 0 when x = −3 or x = 1. Therefore, f is discontinuous at x = −3 and at x = 1 and continuous at every other real number.

D.L. White (Kent State University) 12 / 12