Section 5.4 Continuity of Functions
¶Subsection 5.4.1 Overview
The elementary limit rules for functions tell us that the limit of an algebraic expression made from arithmetic operations will equal the value of the expression at the point in question, if that value exists. So why bother introducing limits at all if they are the same as function evaluation? The fact of the matter is, they aren't the same thing at all. Recall that for piecewise functions, we can use limits to find the limiting value of a function to the left and to the right of a break point. Function evaluation would only allow us to look at the point itself. Having a function value agree with the limits is a characteristic of a function being continuous. A value for x where a function is not defined is an example of a discontinuity. In this section, we consider the continuity of functions. We learn about removable and infinite discontinuities, which correspond to holes and vertical asymptotes in a graph. We learn to compute limits of functions at these discontinuities by looking at simplified, factored expressions. Sign analysis is used for infinite discontinuities to determine whether the discontinuity corresponds to unbounded positive or negative values.Subsection 5.4.2 Removable and Infinite Discontinuities
The intuitive idea of a continuous function is a function whose graph is connected. Sometimes, this is thought of as being able to draw the graph without lifting the pen. The technical definition of a continuity at a point, say at x=c, involves three parts. First, the limit on the left exists. This means that we can trace the graph on a branch with x<c. Second, the limit on the right exists. This means that we also can trace the graph on a branch with x>c. Third, both limits are equal to f(c). This gives us the connection from the left branch to the right branch through the point. Any time a function has a break, it has a discontinuity at that location. A break can be a simple hole, a jump between values, or an infinite discontinuity associated with a vertical asymptote. Discontinuities might also occur due to limits themselves not existing for any reason. Consider two functions, f(x)=1(x−3)(x+2) and g(x)=x2−4x+3x−3. In both functions, the value of the function is not defined at x=3; f and g are both discontinuous at x=3. Consequently, the corresponding limits limx→3f(x) and limx→3g(x) can not be computed directly using the limit rules for functions. If we look at the graphs of our functions, as shown below, we see that there is something fundamentally different about the behavior around x=3. The function f(x)=1(x−3)(x+2) appears to have a vertical asymptote at x=3. The function g(x)=x2−4x+3x−3 looks continuous, even though we know it has a break at x=3.Example 5.4.3.
The function f(x)=3x2−x−2x−1 has a removable discontinuity at x=1. What is the continuous function equivalent to f(x)?
A polynomial, like \(3x^2-x-2\text{,}\) will have a factor of \(x-1\) if and only if that polynomial has a value of 0 when \(x=1\text{.}\) So we can see if it will cancel a factor by checking \(3(1)^2-(1)-2 = 0\text{.}\) Knowing this factor, we can soon find \(3x^2-x-2 = (x-1)(3x+2)\text{.}\) For all \(x \ne 1\text{,}\) we have
We can only say this for \(x \ne 1\) since the domain of \(f\) is \((-\infty,1) \cup (1,\infty)\text{.}\) That is,
Our function \(f(x)\) has the same graph as \(y=3x+2\) except it has a hole at \(x=1\text{.}\)
Theorem 5.4.4.
A rational function f(x)=p(x)q(x) where p and q are polynomial functions has a domain defined by
Further, p and q will have canceling common factors of the form (x−a) where a is a constant if and only if p(a)=0 and q(a)=0.
Example 5.4.5.
Describe the discontinuities of the function
The discontinuities are determined for a rational function by finding the zeros of the polynomial in the denominator, \(q(x)=x^2+x-6\text{.}\) We solve this by factoring:
There are discontinuities (breaks in the graph) at \(x=-3\) and at \(x=2\text{.}\)
We determine the type of discontinuity by seeing if common factors cancel. The numerator \(p(x)=x^3-5x^2+6x\) can be tested even before factoring. At \(x=-3\text{,}\) we have \(p(-3) = -27-5(9)+6(-3) = -90\) so that \(x+3\) is not going to be a common factor. There must be a vertical asymptote at \(x=-3\text{.}\) At \(x=2\text{,}\) we have \(p(2) = 8-5(4)+6(2) = 0\) so that there will be a common factor that cancels.
Because the new formula has a natural domain \(x \ne -3\text{,}\) the discontinuity at \(x=2\) was removable. The graph has a hole at \(x=2\) and a vertical asymptote at \(x=-3\text{.}\) (Notice the addition of an explicit domain on the last step when we canceled, corresponding to the hole.)
Subsection 5.4.3 Limits at Discontinuities
The limit rules do not apply when substitution would result in division by zero. These precisely occur at points of discontinuity. Suppose a rational function f(x)=p(x)q(x) has p(c)=0 and q(c)=0. Immediate substitution of x=c into f(x) would result in 00, which we have earlier identified as an indeterminate limit form. Because p(c)=0 and q(c)=0, p(x) and q(x) have a common factor x−c. Cancellation of that factor gives f(x) a simplified form, and we can try again to evaluate the limit.Example 5.4.6.
Evaluate limx→2x2−5x+6x2−4.
The formula is defined in terms of elementary arithmetic, so we try to evaluate the expression by substituting \(x=2\text{.}\)
The limit has an indeterminate form. We can factor \(x-2\) from numerator and denominator and rewrite the expression.
Limits use the function to the side of the point in question. In this case, \(f(x)\) uses the same formula on the left and the right of the discontinuity. Because the new formula is continuous, we can use substitution.
Example 5.4.7.
Evaluate limx→−2x2−5x+6x2−4.
In the example above, we already found
Attempting substitution, we find
This is an undefined expression and indicates that \(f(x)\) has an infinite discontinuity.
To find the limit as either \(+\infty\) or \(-\infty\text{,}\) we do sign analysis on the simplified formula. The test intervals are separated by the roots and discontinuities. The roots are at solutions to \(x-3=0\text{;}\) the discontinuities are at solutions to \(x+2=0\text{.}\) We have a root at \(x=3\) and a discontinuity at \(x=-2\text{,}\) illustrated in the number line shown below.
For the limit, we need the signs of the function in each interval bordering the point \(x=-2\text{.}\) The intervals to test are \((-\infty,-2)\) and \((-2,3)\text{.}\)
We could update the number line with these signs.
To the left of \(x=-2\text{,}\) we see that \(f(x) \gt 0\) (positive), so a limit from the left at the vertical asymptote must be
To the right of \(x=-2\text{,}\) we see that \(f(x) \lt 0\) (negative), so a limit from the right at the vertical asymptote must be
On the graph of the function, shown below, we see that the graph is unbounded above (\(+\infty\)) to the left of the vertical asymptote and unbounded below (\(-\infty\)) to the right of the vertical asymptote.
Definition 5.4.8.
A function f has a removable discontinuity at x=c if limx→cf(x) exists (left- and right-limits have same value) and limx→cf(x)≠f(c), either because they are different values or f(c) does not exist.
Definition 5.4.9.
A function f has an infinite discontinuity at x=c if one or both of limx→c−f(x) and limx→c+f(x) is infinite. The graph y=f(x) has a vertical asymptote x=c.
Definition 5.4.10.
A function f has an jump discontinuity at x=c if limx→c−f(x) and limx→c+f(x) both exist but limx→c−f(x)≠limx→c+f(x). The graph y=f(x) has a vertical gap between the branches to the left and to the right of x=c.
Subsection 5.4.4 Continuity on Intervals
Having discussed the continuity of functions at individual points, we introduce the idea of describing continuity on intervals. We want to be able to say that the graph of the function is connected over an entire interval. Recall that a limit of a function limx→cf(x) is defined in terms of sequences xn→c with xn≠c. When thinking about continuity on an interval, we also require that the sequences stay in the interval. We begin with open intervals. An open interval (a,b) is the set {x:a<x<b}. Open intervals have the feature that for every value in the set, say c∈(a,b), there will be a sub-interval (a,c) to the left of the point and another sub-interval (c,b) to the right of the point. In relation to a sequence with xn→c, we can deal with left- and right-limits inside the interval.Definition 5.4.11.
A function f is continuous on the open interval (a,b) if for every c∈(a,b), f is continuous at x=c.
Definition 5.4.12.
A function f is continuous on the closed interval [a,b] if for every c∈(a,b), f is continuous at x=c and limx→a+f(x)=f(a) and limx→b−f(x)=f(b).
Definition 5.4.13.
A function f is continuous on an interval I if for every c∈I and every sequence with values xn∈I, xn≠c, and xn→c, we have f(xn)→f(c).
Subsection 5.4.5 Extreme and Intermediate Value Theorems
There are two important theorems that describe what we know about functions that are continuous on closed intervals. The Extreme Value Theorem guarantees that any function that is continuous on a closed interval has a highest and lowest point within that interval. The Intermediate Value Theorem guarantees that a function that is continuous on a closed interval can not skip over any values between its values at the endpoints. The proofs for both of these theorems require advanced methods not taught at this level. We treat them essentially as axioms, statements that are true without proof.Theorem 5.4.14. Extreme Value Theorem.
Suppose f is a function that is continuous on [a,b]. Then there must exist values cm,cM∈[a,b] so that for any x∈[a,b] we have
The values f(cm) and f(cM) are the minimum and maximum values, respectively, of the function f on [a,b].
Example 5.4.15.
Consider the function defined piecewise as
This function has a non-removable discontinuity at x=0, corresponding to a vertical asymptote. Because the formula has x2 in the denominator (always positive), we have
This function is unbounded on the interval [−1,1] and has no maximum. It does have a minimum at f(0)=0 since that is below the rest of the graph.
Example 5.4.16.
Consider the function defined piecewise as
This function has a removable discontinuities at x=±1, where the limits are 1 but the values are 12. In this case, f is continuous on (−1,1) but not on [−1,1]. The maximum value should have been y=1, but the graph never reaches that value because of the discontinuity. The function does have a minimum value at f(0)=0.
Example 5.4.17.
Consider the function defined piecewise as
This function has a removable discontinuity at x=−1. In this case, f is continuous on (−1,1] but not on [−1,1]. In spite of the discontinuity at x=−1, this function has a maximum value f(−1)=2 because that value is above every other point in the interval.
Theorem 5.4.18. Intermediate Value Theorem.
Suppose f is a function that is continuous on [a,b]. Then for every y between f(a) and f(b), there exists some x∈(a,b) so that f(x)=y.
Example 5.4.19.
Consider the function defined piecewise as
This function has a jump discontinuity at x=0, and is otherwise constant. If we consider the interval [−1,1], the values at the endpoints are f(−1)=−1 and f(1)=1. Except for y=0, the function y=f(x) has no solutions for −1<y<1 because of the jump.
Example 5.4.20.
The function f(x)=x3−x−3 is continuous because it is a polynomial and defined everywhere. Because f(1)=−3 and f(2)=3, we know that f(x) must pass through every y-value between -3 and 3 for at least one value of x in the interval (1,2). In particular, if we are solving f(x)=0, since y=0 is between f(1)=−3 and f(2)=3, we know that there is a solution x bracketed by the interval [1,2].
If we find a smaller interval, then we can know more precisely where the root occurs. In particular, since f(1.6)=−0.504 and f(1.7)=0.213 and y=0 is between those values, the Intermediate Value Theorem guarantees that our continuous function has a root bracketed by the interval [1.6,1.7].
Subsection 5.4.6 Summary
- A function defined by an algebraic formula has discontinuities at every point for which the formula is undefined.
- A rational function is defined as the quotient of two polynomials. Discontinuities of rational functions only occur at the zeros of the denominator. If the numerator and denominator have a zero at the same location x=c, then x−c is a common factor that can be cancelled.
- A limit of the form 00 is indeterminate. For rational functions with a limit of this form, we must factor and simplify to continue. If the limit ultimately exists (as a number), the discontinuity is removable and the limit corresponds to having a hole in the graph.
- A rational function with a limit of the form L0 where L≠0 has an infinite discontinuity. The graph of such a function has a vertical asymptote. The left- and right-side limits have signs that are based on sign analysis of the function in the intervals to the left and right of the point of interest.
- A function is continuous on an interval if it continuous at every point in the interval. If an end point is included in the interval, the function must be one-sided continuous from the side contained in the interval.
- The Extreme Value Theorem guarantees that whenever a function f is continuous on a closed interval [a,b], there are points in the interval where f reaches its maximum and minimum (extreme) values restricted to that interval.
- The Intermediate Value Theorem guarantees that whenever a function f is continuous on a closed interval [a,b], the equation f(x)=y has a solution with a<x<b for any y between f(a) and f(b).
- The Intermediate Value Theorem guarantees that a function can only change sign at its roots or discontinuities.