Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs.
We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.
Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.
Our first function of interest is shown in Figure 2.32. We see that the graph of f ( x ) f ( x ) has a hole at a. In fact, f ( a ) f ( a ) is undefined. At the very least, for f ( x ) f ( x ) to be continuous at a, we need the following condition:
i. f ( a ) is defined. i. f ( a ) is defined.Figure 2.32 The function f ( x ) f ( x ) is not continuous at a because f ( a ) f ( a ) is undefined.
However, as we see in Figure 2.33, this condition alone is insufficient to guarantee continuity at the point a. Although f ( a ) f ( a ) is defined, the function has a gap at a. In this example, the gap exists because lim x → a f ( x ) lim x → a f ( x ) does not exist. We must add another condition for continuity at a—namely,
ii. lim x → a f ( x ) exists. ii. lim x → a f ( x ) exists.Figure 2.33 The function f ( x ) f ( x ) is not continuous at a because lim x → a f ( x ) lim x → a f ( x ) does not exist.
However, as we see in Figure 2.34, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to our list:
iii. lim x → a f ( x ) = f ( a ) . iii. lim x → a f ( x ) = f ( a ) .Figure 2.34 The function f ( x ) f ( x ) is not continuous at a because lim x → a f ( x ) ≠ f ( a ) . lim x → a f ( x ) ≠ f ( a ) .
Now we put our list of conditions together and form a definition of continuity at a point.
A function f ( x ) f ( x ) is continuous at a point a if and only if the following three conditions are satisfied:
A function is discontinuous at a point a if it fails to be continuous at a.
The following procedure can be used to analyze the continuity of a function at a point using this definition.
The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.
Using the definition, determine whether the function f ( x ) = ( x 2 − 4 ) / ( x − 2 ) f ( x ) = ( x 2 − 4 ) / ( x − 2 ) is continuous at x = 2 . x = 2 . Justify the conclusion.
Let’s begin by trying to calculate f ( 2 ) . f ( 2 ) . We can see that f ( 2 ) = 0 / 0 , f ( 2 ) = 0 / 0 , which is undefined. Therefore, f ( x ) = x 2 − 4 x − 2 f ( x ) = x 2 − 4 x − 2 is discontinuous at 2 because f ( 2 ) f ( 2 ) is undefined. The graph of f ( x ) f ( x ) is shown in Figure 2.35.
Using the definition, determine whether the function f ( x ) = < − x 2 + 4 if x ≤ 3 4 x − 8 if x >3 f ( x ) = < − x 2 + 4 if x ≤ 3 4 x − 8 if x >3 is continuous at x = 3 . x = 3 . Justify the conclusion.
Let’s begin by trying to calculate f ( 3 ) . f ( 3 ) .
f ( 3 ) = − ( 3 2 ) + 4 = −5 . f ( 3 ) = − ( 3 2 ) + 4 = −5 .Thus, f ( 3 ) f ( 3 ) is defined. Next, we calculate lim x → 3 f ( x ) . lim x → 3 f ( x ) . To do this, we must compute lim x → 3 − f ( x ) lim x → 3 − f ( x ) and lim x → 3 + f ( x ) : lim x → 3 + f ( x ) :
lim x → 3 − f ( x ) = − ( 3 2 ) + 4 = −5 lim x → 3 − f ( x ) = − ( 3 2 ) + 4 = −5 lim x → 3 + f ( x ) = 4 ( 3 ) − 8 = 4 . lim x → 3 + f ( x ) = 4 ( 3 ) − 8 = 4 .Therefore, lim x → 3 f ( x ) lim x → 3 f ( x ) does not exist. Thus, f ( x ) f ( x ) is not continuous at 3. The graph of f ( x ) f ( x ) is shown in Figure 2.36.
Figure 2.36 The function f ( x ) f ( x ) is not continuous at 3 because lim x → 3 f ( x ) lim x → 3 f ( x ) does not exist.
First, observe that
f ( 0 ) = 1 . f ( 0 ) = 1 . lim x → 0 f ( x ) = lim x → 0 sin x x = 1 . lim x → 0 f ( x ) = lim x → 0 sin x x = 1 .Last, compare f ( 0 ) f ( 0 ) and lim x → 0 f ( x ) . lim x → 0 f ( x ) . We see that
f ( 0 ) = 1 = lim x → 0 f ( x ) . f ( 0 ) = 1 = lim x → 0 f ( x ) .Since all three of the conditions in the definition of continuity are satisfied, f ( x ) f ( x ) is continuous at x = 0 . x = 0 .
Using the definition, determine whether the function f ( x ) = < 2 x + 1 if x < 1 2 if x = 1 − x + 4 if x >1 f ( x ) = < 2 x + 1 if x < 1 2 if x = 1 − x + 4 if x >1 is continuous at x = 1 . x = 1 . If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.
By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.
Polynomials and rational functions are continuous at every point in their domains.
Previously, we showed that if p ( x ) p ( x ) and q ( x ) q ( x ) are polynomials, lim x → a p ( x ) = p ( a ) lim x → a p ( x ) = p ( a ) for every polynomial p ( x ) p ( x ) and lim x → a p ( x ) q ( x ) = p ( a ) q ( a ) lim x → a p ( x ) q ( x ) = p ( a ) q ( a ) as long as q ( a ) ≠ 0 . q ( a ) ≠ 0 . Therefore, polynomials and rational functions are continuous on their domains.
We now apply Continuity of Polynomials and Rational Functions to determine the points at which a given rational function is continuous.
For what values of x is f ( x ) = x + 1 x − 5 f ( x ) = x + 1 x − 5 continuous?
The rational function f ( x ) = x + 1 x − 5 f ( x ) = x + 1 x − 5 is continuous for every value of x except x = 5 . x = 5 .
For what values of x is f ( x ) = 3 x 4 − 4 x 2 f ( x ) = 3 x 4 − 4 x 2 continuous?
As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.
These three discontinuities are formally defined as follows:
If f ( x ) f ( x ) is discontinuous at a, then
In Example 2.26, we showed that f ( x ) = x 2 − 4 x − 2 f ( x ) = x 2 − 4 x − 2 is discontinuous at x = 2 . x = 2 . Classify this discontinuity as removable, jump, or infinite.
To classify the discontinuity at 2 we must evaluate lim x → 2 f ( x ) : lim x → 2 f ( x ) :
lim x → 2 f ( x ) = lim x → 2 x 2 − 4 x − 2 = lim x → 2 ( x − 2 ) ( x + 2 ) x − 2 = lim x → 2 ( x + 2 ) = 4 . lim x → 2 f ( x ) = lim x → 2 x 2 − 4 x − 2 = lim x → 2 ( x − 2 ) ( x + 2 ) x − 2 = lim x → 2 ( x + 2 ) = 4 .
Since f is discontinuous at 2 and lim x → 2 f ( x ) lim x → 2 f ( x ) exists, f has a removable discontinuity at x = 2 . x = 2 .
In Example 2.27, we showed that f ( x ) = < − x 2 + 4 if x ≤ 3 4 x − 8 if x >3 f ( x ) = < − x 2 + 4 if x ≤ 3 4 x − 8 if x >3 is discontinuous at x = 3 . x = 3 . Classify this discontinuity as removable, jump, or infinite.
Earlier, we showed that f is discontinuous at 3 because lim x → 3 f ( x ) lim x → 3 f ( x ) does not exist. However, since lim x → 3 − f ( x ) = −5 lim x → 3 − f ( x ) = −5 and lim x → 3 + f ( x ) = 4 lim x → 3 + f ( x ) = 4 both exist, we conclude that the function has a jump discontinuity at 3.
Determine whether f ( x ) = x + 2 x + 1 f ( x ) = x + 2 x + 1 is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, jump, or infinite.
The function value f ( −1 ) f ( −1 ) is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that lim x → −1 − x + 2 x + 1 = − ∞ lim x → −1 − x + 2 x + 1 = − ∞ and lim x → −1 + x + 2 x + 1 = + ∞ . lim x → −1 + x + 2 x + 1 = + ∞ . Therefore, the function has an infinite discontinuity at −1.
For f ( x ) = < x 2 if x ≠ 1 3 if x = 1 , f ( x ) = < x 2 if x ≠ 1 3 if x = 1 , decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.
Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval . As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.
A function f ( x ) f ( x ) is said to be continuous from the right at a if lim x → a + f ( x ) = f ( a ) . lim x → a + f ( x ) = f ( a ) .
A function f ( x ) f ( x ) is said to be continuous from the left at a if lim x → a − f ( x ) = f ( a ) . lim x → a − f ( x ) = f ( a ) .
A function is continuous over an open interval if it is continuous at every point in the interval. A function f ( x ) f ( x ) is continuous over a closed interval of the form [ a , b ] [ a , b ] if it is continuous at every point in ( a , b ) ( a , b ) and is continuous from the right at a and is continuous from the left at b. Analogously, a function f ( x ) f ( x ) is continuous over an interval of the form ( a , b ] ( a , b ] if it is continuous over ( a , b ) ( a , b ) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.
Requiring that lim x → a + f ( x ) = f ( a ) lim x → a + f ( x ) = f ( a ) and lim x → b − f ( x ) = f ( b ) lim x → b − f ( x ) = f ( b ) ensures that we can trace the graph of the function from the point ( a , f ( a ) ) ( a , f ( a ) ) to the point ( b , f ( b ) ) ( b , f ( b ) ) without lifting the pencil. If, for example, lim x → a + f ( x ) ≠ f ( a ) , lim x → a + f ( x ) ≠ f ( a ) , we would need to lift our pencil to jump from f ( a ) f ( a ) to the graph of the rest of the function over ( a , b ] . ( a , b ] .
State the interval(s) over which the function f ( x ) = x − 1 x 2 + 2 x f ( x ) = x − 1 x 2 + 2 x is continuous.
Since f ( x ) = x − 1 x 2 + 2 x f ( x ) = x − 1 x 2 + 2 x is a rational function, it is continuous at every point in its domain. The domain of f ( x ) f ( x ) is the set ( − ∞ , −2 ) ∪ ( −2 , 0 ) ∪ ( 0 , + ∞ ) . ( − ∞ , −2 ) ∪ ( −2 , 0 ) ∪ ( 0 , + ∞ ) . Thus, f ( x ) f ( x ) is continuous over each of the intervals ( − ∞ , −2 ) , ( −2 , 0 ) , ( − ∞ , −2 ) , ( −2 , 0 ) , and ( 0 , + ∞ ) . ( 0 , + ∞ ) .
State the interval(s) over which the function f ( x ) = 4 − x 2 f ( x ) = 4 − x 2 is continuous.
From the limit laws, we know that lim x → a 4 − x 2 = 4 − a 2 lim x → a 4 − x 2 = 4 − a 2 for all values of a in ( −2 , 2 ) . ( −2 , 2 ) . We also know that lim x → −2 + 4 − x 2 = 0 lim x → −2 + 4 − x 2 = 0 exists and lim x → 2 − 4 − x 2 = 0 lim x → 2 − 4 − x 2 = 0 exists. Therefore, f ( x ) f ( x ) is continuous over the interval [ −2 , 2 ] . [ −2 , 2 ] .
State the interval(s) over which the function f ( x ) = x + 3 f ( x ) = x + 3 is continuous.
The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.
If f ( x ) f ( x ) is continuous at L and lim x → a g ( x ) = L , lim x → a g ( x ) = L , then
lim x → a f ( g ( x ) ) = f ( lim x → a g ( x ) ) = f ( L ) . lim x → a f ( g ( x ) ) = f ( lim x → a g ( x ) ) = f ( L ) .
Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed lim x → 0 cos x = 1 = cos ( 0 ) . lim x → 0 cos x = 1 = cos ( 0 ) . Consequently, we know that f ( x ) = cos x f ( x ) = cos x is continuous at 0. In Example 2.35 we see how to combine this result with the composite function theorem.
Evaluate lim x → π / 2 cos ( x − π 2 ) . lim x → π / 2 cos ( x − π 2 ) .
The given function is a composite of cos x cos x and x − π 2 . x − π 2 . Since lim x → π / 2 ( x − π 2 ) = 0 lim x → π / 2 ( x − π 2 ) = 0 and cos x cos x is continuous at 0, we may apply the composite function theorem. Thus,
lim x → π / 2 cos ( x − π 2 ) = cos ( lim x → π / 2 ( x − π 2 ) ) = cos ( 0 ) = 1 . lim x → π / 2 cos ( x − π 2 ) = cos ( lim x → π / 2 ( x − π 2 ) ) = cos ( 0 ) = 1 .
Evaluate lim x → π sin ( x − π ) . lim x → π sin ( x − π ) .
The proof of the next theorem uses the composite function theorem as well as the continuity of f ( x ) = sin x f ( x ) = sin x and g ( x ) = cos x g ( x ) = cos x at the point 0 to show that trigonometric functions are continuous over their entire domains.
Trigonometric functions are continuous over their entire domains.
We begin by demonstrating that cos x cos x is continuous at every real number. To do this, we must show that lim x → a cos x = cos a lim x → a cos x = cos a for all values of a.
lim x → a cos x = lim x → a cos ( ( x − a ) + a ) rewrite x = x − a + a = lim x → a ( cos ( x − a ) cos a − sin ( x − a ) sin a ) apply the identity for the cosine of the sum of two angles = cos ( lim x → a ( x − a ) ) cos a − sin ( lim x → a ( x − a ) ) sin a lim x → a ( x − a ) = 0 , and sin x and cos x are continuous at 0 = cos ( 0 ) cos a − sin ( 0 ) sin a evaluate cos(0) and sin(0) and simplify = 1 · cos a − 0 · sin a = cos a . lim x → a cos x = lim x → a cos ( ( x − a ) + a ) rewrite x = x − a + a = lim x → a ( cos ( x − a ) cos a − sin ( x − a ) sin a ) apply the identity for the cosine of the sum of two angles = cos ( lim x → a ( x − a ) ) cos a − sin ( lim x → a ( x − a ) ) sin a lim x → a ( x − a ) = 0 , and sin x and cos x are continuous at 0 = cos ( 0 ) cos a − sin ( 0 ) sin a evaluate cos(0) and sin(0) and simplify = 1 · cos a − 0 · sin a = cos a .
The proof that sin x sin x is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of sin x sin x and cos x , cos x , their continuity follows from the quotient limit law.
As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.
Functions that are continuous over intervals of the form [ a , b ] , [ a , b ] , where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem .
Let f be continuous over a closed, bounded interval [ a , b ] . [ a , b ] . If z is any real number between f ( a ) f ( a ) and f ( b ) , f ( b ) , then there is a number c in [ a , b ] [ a , b ] satisfying f ( c ) = z f ( c ) = z in Figure 2.38.
