1.8 Continuity

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In this section we learn the definition of continuity and we study the types of discontinuities.

Continuity

One of the tools we used in the last section for evaluating limits analytically was plugging in. For example, to compute $\lim _{x \to 5} x^2$ we would simply plug in $x = 5$ and get the correct answer of 25. In most of the examples in the last section, plugging in did not give an acceptable answer at first, so more work had to be done before eventually plugging in. If plugging does work at the outset, then the function is called continuous.

Continuity The function $f(x)$ is continuous at $x = a$ if: $\lim _{x \to a} f(x) = f(a).$

Since $\lim _{x \to 5} x^2 = 5^2 = 25,$ we can say that the function $f(x) = x^2$ is continuous at $x = 5$ .

Since $\lim _{x \to 4} \sqrt {x} = \sqrt {4} = 2,$

we can say that the function
$f(x) = \answer {\sqrt x}$
is continuous at
$x = \answer {4}.$

Since $\lim _{x \to \pi } \cos (x) = \cos (\pi ) = -1,$ we can say that the function $f(x) = \cos (x)$ is continuous at $x = \pi$ .

Since $\lim _{x \to 3} 2^x = 2^3 = 8,$

we can say that the function
$f(x) = \answer {2^x}$
is continuous at
$x = \answer {3}.$

Since $\lim _{x \to 1} \ln (x) = \ln (1) = 0,$ we can say that the function $f(x) = \ln (x)$ is continuous at $x = 1$ .

Since $\lim _{x \to \sqrt 3} \tan ^{-1}(x) = \tan ^{-1}(\sqrt 3) = \pi /3,$

we can say that the function
$f(x) = \answer {\tan ^{-1}(x)}$
is continuous at
$x = \answer {\sqrt 3}.$

If a function is continuous at each point in an interval, $I$ , then we say that $f(x)$ is continuous on $I$ .

A function $f(x)$ is called continuous from the left at $x=a$ if $\lim _{x \to a^-} f(x) = f(a),$ and it is called continuous from the right at $x = a$ if $\lim _{x \to a^+} f(x) = f(a).$ Note that if $f(x)$ is continuous at $x=a$ then it is both continuous from the left and continuous from the right at $x = a$ .

If for some reason, a limit cannot be computed by plugging in, then we say that the function is discontinuous. In other words, if $\lim _{x \to a} f(x) \neq f(a),$ then we say that $f(x)$ has a discontinuity at $x = a$ . Next, we explore the types of discontinuities.

Types of Discontinuities

A function $f(x)$ has a discontinuity at $x = a$ if $\lim _{x \to a} f(x) \neq f(a).$ There are four main types of discontinuities: removable, jump, infinite and essential.

First, a discontinuity is called a removable discontinuity if $\lim _{x \to a} f(x) \ \ \text {exists and is finite }.$

Here are two examples of graphs of functions that have removable discontinuities:

The second type of discontinuity is called a jump discontinuity. This happens when the one-sided limits are different numbers, so that

$\lim _{x \to a^-} f(x) \neq \lim _{x \to a^+} f(x).$ and hence $\lim _{x \to a} f(x) \ \text {DNE}.$

It is important in this case that the one-sided limits are both finite.

As an example, the function $f(x) = \frac {|x|}{x}$ has a jump discontinuity at $x = 0$ because $\lim _{x \to 0^-} \frac {|x|}{x} = -1 \text { but}$ $\lim _{x \to 0^+} \frac {|x|}{x} = 1.$

A graph of this function is shown below.

Here are two examples of graphs of functions that have jump discontinuities:

The third type of discontinuity is called an infinite discontinuity. This occurs when either of the one-sided limits is either $\pm \infty$ , i.e., $\lim _{x \to a^-} f(x) = \pm \infty \ \text {or}$ $\lim _{x \to a^+} f(x) = \pm \infty .$ The graph of the function $f(x)$ has a vertical asymptote at an infinite discontinuity. The function $f(x) = 1/x$ , has an infinite discontinuity at $x=0$ since $\lim _{x \to 0^-} \frac {1}{x} = -\infty \ \text { and}$ $\lim _{x \to 0^+} \frac {1}{x} = \infty .$ The graph of $f(x) = 1/x$ is shown below. $\graph {1/x}$

As another example, the function $f(x) = \tan (x)$ has an infinite discontinuity at $x = \frac {\pi }{2}$ since $\lim _{x \to \frac {\pi }{2}^-} \tan (x) = \infty \ \text { and}$ $\lim _{x \to \frac {\pi }{2}^+} \tan (x) = -\infty .$ In fact, $f(x) = \tan (x)$ has infinite discontinuities at all odd multiples of $\pi /2$ . The graph of $f(x) = \tan (x)$ is below.

$\graph {tan(x)}$

Finally, if a discontinuity is not one of the first three types, it is called an essential discontinuity. As an example, the function $f(x) = \sin (\frac {1}{x})$ , shown below, has an essential discontinuity at $x = 0$ . Neither of one-sided limits at $x=0$ exist due to oscillation of the function, and the function does not have a vertical asymptote since $-1 \leq \sin \big (\frac {1}{x}\big ) \leq 1$ for all values of $x$ except $x = 0$ , where the function is undefined.

$\graph {sin(1/x)}$

Which type of discontinuity does $f(x)$ have at $x=2$ if $\lim _{x \to 2^+} f(x) = -\infty ?$

removable jump infinite essential

Which type of discontinuity does $f(x)$ have at $x=2$ if $\lim _{x \to 2^-} f(x) = \lim _{x \to 2^+} f(x),$ but $f(2)$ is undefined?

removable jump infinite essential

Which type of discontinuity does $f(x)$ have at $x=2$ if $\lim _{x \to 2^+} f(x) \ \text {DNE}?$

removable jump infinite essential

Which type of discontinuity does $f(x)$ have at $x=2$ if $\lim _{x \to 2^-} f(x) = 3$ and $\lim _{x \to 2^+} f(x) = -1$

removable jump infinite essential

Piecewise Functions

In this section we will examine the continuity of piecewise defined functions.

Is the function defined below continuous at $x = 2$ ? $f(x) = \left \{ \begin {array}{lr} x-1 & \ \text {for} \ x \leq 2 \\ 3 & \ \text {for} \ x > 2 \end {array} \right .$

We need to compute $f(2)$ and the one-sided limits as $x$ approaches 2. If all three of these are equal, then $f(x)$ is continuous at $x=2$ . Otherwise, $f(x)$ has a discontinuity at $x=2$ .
First, $f(2) = 2-1 = 1.$ Next, $\lim _{x \to 2^-} f(x) = \lim _{x \to 2^-} x-1 = 2-1 = 1.$ Finally, $\lim _{x \to 2^+} f(x) = \lim _{x \to 2^+}3 = 3.$ Since the one-sided limits are different, the two-sided limit $\lim _{x \to 2} f(x) \ \text {DNE}$ and hence $f(x)$ is not continuous at $x = 2$ . In this case, $f(x)$ has a jump discontinuity at $x=2$ since the one-sided limits are finite but different.

Determine whether the function $f(x)$ given below is continuous at $x = 1$ . $f(x) = \left \{ \begin {array}{lr} 2x & \ \text {for} \ x < 1 \\ x^2 + 2 & \ \text {for} \ x \geq 1 \end {array} \right .$ $f(1) = \answer {3}$ $\lim _{x \to 1^-} f(x) = \answer {2}$ $\lim _{x \to 1^+} f(x) = \answer {3}$ Is $f(x)$ continuous at $x = 1$ ?

Yes No

Is the function define below continuous at $x = 2$ ? $f(x) = \left \{ \begin {array}{lr} x-1 & \ \text {for} \ x \leq 2 \\ x^2 - 3 & \ \text {for} \ x > 2 \end {array} \right .$

We need to compute $f(2)$ and the one-sided limits as $x$ approaches 2. If all three of these are equal, then $f(x)$ is continuous at $x=2$ . Otherwise, $f(x)$ has a discontinuity at $x=2$ .
First, $f(2) = 2-1 = 1.$ Next, $\lim _{x \to 2^-} f(x) = \lim _{x \to 2^-} x-1 = 2-1 = 1,$ and finally, $\lim _{x \to 2^+} f(x) = \lim _{x \to 2^+} x^2 - 3 = 2^2 - 3 = 1.$ Since all three of these values are equal, we can write $\lim _{x \to 2} f(x) = f(2)$ and conclude that $f(x)$ is continuous at $x = 2$ .

Determine whether the function $f(x)$ given below is continuous at $x = 1$ . $f(x) = \left \{ \begin {array}{lr} 2x & \ \text {for} \ x < 1 \\ x^2 + 1 & \ \text {for} \ x \geq 1 \end {array} \right .$ $f(1) = \answer {2}$ $\lim _{x \to 1^-} f(x) = \answer {2}$ $\lim _{x \to 1^+} f(x) = \answer {2}$ Is $f(x)$ continuous at $x = 1$ ?

Yes No

Is the function $f(x)$ defined below continuous at $x = 2$ ? $f(x) = \left \{ \begin {array}{lr} x-1 & \ \text {for} \ x < 2 \\ 3 & \ \text {for} \ x = 2 \\ x^2 - 3 & \ \text {for} \ x > 2 \end {array} \right .$

We need to compute $f(2)$ and the one-sided limits as $x$ approaches 2. If all three of these are equal, then $f(x)$ is continuous at $x=2$ . Otherwise, $f(x)$ has a discontinuity at $x=2$ .
First, $f(2) = 3 \ \text {(given)}.$ Next, $\lim _{x \to 2^-} f(x) = \lim _{x \to 2^-} x-1 = 2-1 = 1$ and $\lim _{x \to 2^+} f(x) = \lim _{x \to 2^+} x^2 - 3 = 2^2 - 3 = 1.$ Since the one-sided limits are equal, the two-sided limit exists: $\lim _{x \to 2} f(x) = 1.$ However, $\lim _{x \to 2} f(x) \neq f(2) = 3.$ Hence $f(x)$ is not continuous at $x = 2$ . This is an example of a removable discontinuity, since $\lim _{x \to 2} f(x)$ is a finite number.

Determine whether the function $f(x)$ given below is continuous at $x = 1$ . $f(x) = \left \{ \begin {array}{lr} 2x & \ \text {for} \ x < 1 \\ 5 & \ \text {for} \ x = 1 \\ x^2 + 1 & \ \text {for} \ x > 1 \end {array} \right .$ $f(1) = \answer {5}$ $\lim _{x \to 1^-} f(x) = \answer {2}$ $\lim _{x \to 1^+} f(x) = \answer {2}$ Is $f(x)$ continuous at $x = 1$ ?

Yes No

Determine the value of $k$ that will make the function $f(x)$ given below continuous at $x = 2$ .

$f(x) = \left \{ \begin {array}{lr} kx-5 & \ \text {for} \ x \leq 2 \\ x^2 - k & \ \text {for} \ x > 2 \end {array} \right .$

We need to compute $f(2)$ and the one-sided limits as $x$ approaches 2. We then set these expressions equal to each other and solve for $k$ .
First, $f(2) = 2k-5.$ Next, $\lim _{x \to 2^-} f(x) = \lim _{x \to 2^-} kx-5 = 2k-5,$ and $\lim _{x \to 2^+} f(x) = \lim _{x \to 2^+} x^2 - k = 2^2 - k = 4 - k.$ Setting these expressions equal to one another gives the equation: $2k - 5 = 4 - k.$ Solving for $k$ gives $3k = 9$ and hence, $k = 3.$ This value of $k$ will make the function $f(x)$ continuous at $x=2$ .

Determine the value of $k$ that will make the function $f(x)$ given below continuous at $x = 1$ .

$f(x) = \left \{ \begin {array}{lr} x^2 - k & \ \text {for} \ x < 1 \\ kx-3 & \ \text {for} \ x \geq 1 \end {array} \right .$

$f(1) = \answer {k-3}$ $\lim _{x \to 1^-} f(x) = \answer {1-k}$ $\lim _{x \to 1^+} f(x) = \answer {k-3}$ The value of $k$ that makes $f(x)$ continuous at $x = 1$ is $k = \answer {2}$

Continuity of Familiar Functions

Consider the function $f(x) = x^2$ and any number $a$ . We can compute the limit $\lim _{x \to a} x^2 = a^2$ by plugging in. Therefore, we can say that $f(x) = x^2$ is continuous for all real numbers. We can also say that $f(x) = x^2$ is continuous on the interval $(-\infty , \infty )$ . Actually, this is true for all polynomials, including constant functions. If $p(x)$ is a polynomial, then $p(x)$ is continuous on the interval $(-\infty , \infty )$ .
There are some other familiar functions which are also continuous on the interval $(-\infty , \infty )$ . These are: $f(x) = \sin (x), \cos (x), e^x, \tan ^{-1}(x), \sqrt [3] x \ \text {and} \ |x|.$ The function $f(x) = \ln (x)$ is only defined for $x$ in the interval $(0, \infty )$ and it is continuous on this interval.
The function $f(x) = \tan (x)$ has vertical asymptotes at odd multiples of $\frac {\pi }{2}$ . It is continuous between these vertical asymptotes, so, for example, $f(x) = \tan (x)$ is continuous on the interval $(-\frac {\pi }{2},\frac {\pi }{2})$ . The function $f(x) = \sec (x)$ is similar to $\tan (x)$ in that it has vertical asymptotes at odd multiples of $\frac {\pi }{2}$ , and it is continuous on the intervals between them. The functions $\cot (x)$ and $\csc (x)$ have vertical asymptotes at multiples of $\pi$ and like $\tan (x)$ and $\sec (x)$ , they are continuous between their asymptotes. For example, the function $f(x) = \cot (x)$ is continuous on the interval $(0, \pi )$ .

The function $f(x) = \sqrt x$ is only defined for $x \geq 0$ and it is continuous for all of these values of $x$ . In other words, $f(x) = \sqrt x$ is continuous on the interval $[0, \infty )$ . To say that the function is continuous at the left endpoint of this interval ( $x = 0$ ) it is sufficient that the function is right continuous at this point. And it is indeed true that $\lim _{x \to 0^+} \sqrt x = \sqrt 0 = 0.$ In general, a root function, $f(x) = \sqrt [n] x$ is continuous on the interval $[0, \infty )$ if $n$ is even and it is continuous on the interval $(-\infty , \infty )$ if $n$ is odd.

The last type of familiar function that we will discuss here is the rational function. A rational function is a ratio of polynomials, $f(x) = \frac {p_1(x)}{p_2(x)},$ where $p_1(x)$ and $p_2(x)$ are both polynomials and the degree of $p_2(x)$ is at least 1. Such a function is continuous for all values of $x$ such that $p_2(x) \neq 0$ . For example, the function $f(x) = \frac {x}{x^2 + 1}$ is continuous on the interval $(-\infty , \infty )$ since $x^2 + 1 \neq 0$ for any $x$ . On the other hand, the function $f(x) = \frac {x}{x^2 - 1}$ is continuous on the intervals $(-\infty , -1), (-1, 1)$ and $(1, \infty )$ since $x^2 - 1 = 0$ when $x = \pm 1$ .

Properties of Continuity

Continuous functions combine nicely with respect to the operations addition, subtraction, multiplication, division and composition. Specifically, if $f(x)$ and $g(x)$ are both continuous at $x = a$ , then so are $f(x) + g(x),$ $f(x) - g(x) \ \text {and}$ $f(x) \cdot g(x).$ Furthermore, if $g(a) \neq 0$ then $\frac {f(x)}{g(x)}$ is also continuous at $x = a$ . In words, we say that the sum, difference, product and quotient of continuous functions is continuous (with the understanding that $g(a) \neq 0$ in the case of the quotient.) Things are slightly more complicated for the composition. If $g(x)$ is continuous at $x = a$ and $f(x)$ is continuous at $x = g(a)$ then then composition $f(g(x))$ is continuous at $x = a$ . This situation is different from the four basic operations because in composition, when plugging in $x = a$ , we plug $a$ into $g(x)$ and then plug $g(a)$ into $f(x)$ , whereas for the first four basic operations we plug $x = a$ into both $f(x)$ and $g(x)$ .

Here is a detailed, lecture style video on discontinuities: