Continuity

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.

Continuity at a 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 [link]. We see that the graph of

f (x)

However, as we see in [link], this condition alone is insufficient to guarantee continuity at the point a. Although

f (a)

is defined, the function has a gap at a. In this example, the gap exists because

lim_{x \to a} f (x)

However, as we see in [link], 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:

Now we put our list of conditions together and form a definition of continuity at a point.

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.

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Proof

We now apply [link] to determine the points at which a given rational function is continuous.

Types of Discontinuities

As we have seen in [link] and [link], 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. [link] 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.

Continuity over an Interval

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 is continuous over an open interval if it is continuous at every point in the interval. A function

f (x)

and is continuous from the right at a and is continuous from the left at b. Analogously, a function

f (x)

and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.

without lifting the pencil. If, for example,

lim_{x \to a^{+}} f (x) \neq f (a),

The [link] 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.

Before we move on to [link], recall that earlier, in the section on limit laws, we showed

lim_{x \to 0} cos x = 1 = cos (0) .

is continuous at 0. In [link] we see how to combine this result with the composite function theorem.

The proof of the next theorem uses the composite function theorem as well as the continuity of

f (x) = sin x

at the point 0 to show that trigonometric functions are continuous over their entire domains.

Proof

is continuous at every real number. To do this, we must show that

lim_{x \to a} cos x = cos a

is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of

sin x

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.

The Intermediate Value Theorem

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.

Key Concepts

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

f (x) = \frac{1}{\sqrt{x}}

The function is defined for all x in the interval $(0, \infty) .$

f (x) = \frac{2}{x^{2} + 1}

f (x) = \frac{x}{x^{2} - x}

Removable discontinuity at $x = 0;$

infinite discontinuity at $x = 1$

g (t) = t^{−1} + 1

f (x) = \frac{5}{e^{x} - 2}

Infinite discontinuity at $x = ln 2$

f (x) = \frac{\| x - 2 \|}{x - 2}

H (x) = tan 2 x

Infinite discontinuities at $x = \frac{(2 k + 1) π}{4},$

for $k = 0, \pm 1, \pm 2, \pm 3,…$

f (t) = \frac{t + 3}{t^{2} + 5 t + 6}

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

\frac{2 x^{2} - 5 x + 3}{x - 1}

at $x = 1$

No. It is a removable discontinuity.

h (θ) = \frac{sin θ - cos θ}{tan θ}

at $θ = π$

g (u) = {\begin{cases} \frac{6 u^{2} + u - 2}{2 u - 1} & if u \neq \frac{1}{2} \\ \frac{7}{2} & if u = \frac{1}{2} \end{cases},

at $u = \frac{1}{2}$

Yes. It is continuous.

f (y) = \frac{sin (π y)}{tan (π y)},

at $y = 1$

f (x) = {\begin{cases} x^{2} - e^{x} & if x < 0 \\ x - 1 & if x \geq 0 \end{cases},

at $x = 0$

Yes. It is continuous.

f (x) = {\begin{cases} x sin (x) if x \leq π \\ x tan (x) if x > π \end{cases},

at $x = π$

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

f (x) = {\begin{matrix} 3 x + 2, & x < k \\ 2 x - 3, & k \leq x \leq 8 \end{matrix}

k = −5

f (θ) = {\begin{cases} sin θ, & 0 \leq θ < \frac{π}{2} \\ cos (θ + k), & \frac{π}{2} \leq θ \leq π \end{cases}

f (x) = {\begin{matrix} \frac{x^{2} + 3 x + 2}{x + 2}, & x \neq - 2 \\ k, & x = −2 \end{matrix}

k = −1

f (x) = {\begin{matrix} e^{k x}, & 0 \leq x < 4 \\ x + 3, & 4 \leq x \leq 8 \end{matrix}

f (x) = {\begin{matrix} \sqrt{k x}, & 0 \leq x \leq 3 \\ x + 1, & 3 < x \leq 10 \end{matrix}

k = \frac{16}{3}

In the following exercises, use the Intermediate Value Theorem (IVT).

Let $h (x) = {\begin{cases} 3 x^{2} - 4, & x \leq 2 \\ 5 + 4 x, & x > 2 \end{cases}$

Over the interval $[0, 4],$

there is no value of x such that $h (x) = 10,$

although $h (0) < 10$

and $h (4) > 10 .$

Explain why this does not contradict the IVT.

A particle moving along a line has at each time t a position function $s (t),$

which is continuous. Assume $s (2) = 5$

and $s (5) = 2 .$

Another particle moves such that its position is given by $h (t) = s (t) - t .$

Explain why there must be a value c for $2 < c < 5$

such that $h (c) = 0 .$

Since both s and $y = t$

are continuous everywhere, then $h (t) = s (t) - t$

is continuous everywhere and, in particular, it is continuous over the closed interval $[2, 5] .$

Also, $h (2) = 3 > 0$

and $h (5) = −3 < 0 .$

Therefore, by the IVT, there is a value $x = c$

such that $h (c) = 0 .$

[T] Use the statement “The cosine of t is equal to t cubed.”

Write a mathematical equation of the statement.
Prove that the equation in part a. has at least one real solution.
Use a calculator to find an interval of length 0.01 that contains a solution.

Apply the IVT to determine whether $2^{x} = x^{3}$

has a solution in one of the intervals $[1.25, 1.375]$

or $[1.375, 1.5] .$

Briefly explain your response for each interval.

The function $f (x) = 2^{x} - x^{3}$

is continuous over the interval $[1.25, 1.375]$

and has opposite signs at the endpoints.

Consider the graph of the function $y = f (x)$

shown in the following graph.

Find all values for which the function is discontinuous.
For each value in part a., state why the formal definition of continuity does not apply.
Classify each discontinuity as either jump, removable, or infinite.

Let $f (x) = {\begin{matrix} 3 x, x > 1 \\ x^{3}, x < 1 \end{matrix} .$

Sketch the graph of f.
Is it possible to find a value k such that $f (1) = k,$
which makes
$f (x)$
continuous for all real numbers? Briefly explain.

a.* * *

A graph of the given piecewise function containing two segments. The first, x^3, exists for x < 1 and ends with an open circle at (1,1). The second, 3x, exists for x > 1. It beings with an open circle at (1,3).

b. It is not possible to redefine $f (1)$

since the discontinuity is a jump discontinuity.

Let $f (x) = \frac{x^{4} - 1}{x^{2} - 1}$

for $x \neq - 1, 1 .$

Sketch the graph of f.
Is it possible to find values $k_{1}$
and
$k_{2}$
such that
$f (−1) = k$
and
$f (1) = k_{2},$
and that makes
$f (x)$
continuous for all real numbers? Briefly explain.

Sketch the graph of the function $y = f (x)$

with properties i. through vii.

The domain of f is $(− \infty, + \infty) .$
f has an infinite discontinuity at $x = −6 .$
$f (−6) = 3$
$lim_{x \to {−3}^{-}} f (x) = lim_{x \to {−3}^{+}} f (x) = 2$
$f (−3) = 3$
f is left continuous but not right continuous at $x = 3 .$
$lim_{x \to - \infty} f (x) = − \infty$
and
$lim_{x \to + \infty} f (x) = + \infty$

Answers may vary; see the following example:* * *

Sketch the graph of the function $y = f (x)$

with properties i. through iv.

The domain of f is $[0, 5] .$
$lim_{x \to 1^{+}} f (x)$
and
$lim_{x \to 1^{-}} f (x)$
exist and are equal.
$f (x)$
is left continuous but not continuous at
$x = 2,$
and right continuous but not continuous at
$x = 3 .$
$f (x)$
has a removable discontinuity at
$x = 1,$
a jump discontinuity at
$x = 2,$
and the following limits hold:
$lim_{x \to 3^{-}} f (x) = − \infty$
and
$lim_{x \to 3^{+}} f (x) = 2 .$

In the following exercises, suppose $y = f (x)$

is defined for all x. For each description, sketch a graph with the indicated property.

Discontinuous at $x = 1$

with $lim_{x \to −1} f (x) = −1$

and $lim_{x \to 2} f (x) = 4$

Answers may vary; see the following example:* * *

The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x < 1. It ends at (1,1). The second part is an increasing line that exists for x > 1. It begins at (1,3).

Discontinuous at $x = 2$

but continuous elsewhere with $lim_{x \to 0} f (x) = \frac{1}{2}$

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

f (t) = \frac{2}{e^{t} - e^{- t}}

is continuous everywhere.

False. It is continuous over $(− \infty, 0) \cup (0, \infty) .$

If the left- and right-hand limits of $f (x)$

as $x \to a$

exist and are equal, then f cannot be discontinuous at $x = a .$

If a function is not continuous at a point, then it is not defined at that point.

False. Consider $f (x) = {\begin{cases} x if x \neq 0 \\ 4 if x = 0 \end{cases} .$

According to the IVT, $cos x - sin x - x = 2$

has a solution over the interval $[−1, 1] .$

If $f (x)$

is continuous such that $f (a)$

and $f (b)$

have opposite signs, then $f (x) = 0$

has exactly one solution in $[a, b] .$

False. Consider $f (x) = cos (x)$

on $[- π, 2 π] .$

The function $f (x) = \frac{x^{2} - 4 x + 3}{x^{2} - 1}$

is continuous over the interval $[0, 3] .$

If $f (x)$

is continuous everywhere and $f (a), f (b) > 0,$

then there is no root of $f (x)$

in the interval $[a, b] .$

False. The IVT does not work in reverse! Consider ${(x - 1)}^{2}$

over the interval $[−2, 2] .$

[T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation $F (r) = k_{e} \frac{\| q_{1} q_{2} \|}{r^{2}},$

where $k_{e}$

is Coulomb’s constant, $q_{i}$

are the magnitudes of the charges of the two particles, and r is the distance between the two particles.

To simplify the calculation of a model with many interacting particles, after some threshold value $r = R,$

we approximate F as zero.

Explain the physical reasoning behind this assumption.
What is the force equation?
Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of $1.6022 \times 10^{−19} coulombs (C),$
and the Coulomb constant
$k_{e} = 8.988 \times 10^{9} {Nm}^{2} / C^{2}$
are 1 m apart. Also, assume
$R < 1 m .$
How much inaccuracy does our approximation generate? Is our approximation reasonable?
Is there any finite value of R for which this system remains continuous at R?

Instead of making the force 0 at R, instead we let the force be 10⁻²⁰ for $r \geq R .$

Assume two protons, which have a magnitude of charge $1.6022 \times 10^{−19} C,$

and the Coulomb constant $k_{e} = 8.988 \times 10^{9} {Nm}^{2} / C^{2} .$

Is there a value R that can make this system continuous? If so, find it.

R = 0.0001519 m

Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by $F (d) = - m k / d^{2},$

where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant.

[T] Determine the value and units of k given that the mass of the rocket on Earth is 3 million kg. (Hint: The distance from the center of Earth to its surface is 6378 km.)

[T] After a certain distance D has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by $F (d) = {\begin{cases} - \frac{m k}{d^{2}} & if d < D \\ 10,000 & if d \geq D \end{cases} .$

Find the necessary condition D such that the force function remains continuous.

D = 63.78 km

As the rocket travels away from Earth’s surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as $F (d) = {\begin{cases} - \frac{m_{1} k}{d^{2}} if d < D \\ - \frac{m_{2} k}{d^{2}} if d \geq D \end{cases} .$

Is there a D value such that this function is continuous, assuming $m_{1} \neq m_{2} ?$

Prove the following functions are continuous everywhere

f (θ) = sin θ

For all values of $a, f (a)$

is defined, $lim_{θ \to a} f (θ)$

exists, and $lim_{θ \to a} f (θ) = f (a) .$

Therefore, $f (θ)$

is continuous everywhere.

g (x) = \| x \|

Where is $f (x) = {\begin{cases} 0 if x is irrational \\ 1 if x is rational \end{cases}$

continuous?

Nowhere

Continuity at a Point

Proof

Types of Discontinuities

Continuity over an Interval

Proof

The Intermediate Value Theorem

Key Concepts

Glossary