Continuity

On a graph, continuity at

a

means the curve arrives at the point

(a, f(a))

from both sides with no break, although it may turn a corner there.

Types of discontinuity

Removable: the limit at 2 is 2, but the value there is 3.5.

Jump: the one-sided limits at 2 are 3 and 1.

Infinite: the one-sided limits at 2 are −∞ and ∞.

A removable discontinuity is called that because defining or redefining one value,

f(a)

, would make

f

continuous at

a

. No single value can fix a jump or an infinite discontinuity. Other kinds exist, such as the oscillation of

\sin\frac{1}{x}

0

, but these three are the ones to classify.

Example 1

Let

k(x) = \frac{x^2 - 1}{x^3 - x}.

Find each

x

where

k

is discontinuous, and classify each discontinuity.

The denominator is

x(x - 1)(x + 1)

, which is

0

x = 0

1

, and

-1

. The function isn't defined at any of them, so it's discontinuous at all three. The numerator factors as

(x - 1)(x + 1)

, so for every other

x

k(x) = \frac{(x - 1)(x + 1)}{x(x - 1)(x + 1)} = \frac{1}{x}.

x = 1

and

x = -1

\lim_{x \to 1} k(x) = 1

and

\lim_{x \to -1} k(x) = -1

. The limits exist, but

k

isn't defined there, so both discontinuities are removable.

x = 0

\lim_{x \to 0^-} k(x) = -\infty

and

\lim_{x \to 0^+} k(x) = \infty

, so the discontinuity is infinite.

The graph of k: holes at (1, 1) and (−1, −1), and the y-axis as a vertical asymptote.

Example 2

Determine whether

g

is continuous at

x = -2

and at

x = 2

g(x) = \begin{cases} x + 4, & x < -2 \\ x^2 - 2, & -2 \le x \le 2 \\ 7 - x, & x > 2 \end{cases}

x = -2

g(-2) = (-2)^2 - 2 = 2

, and

\begin{aligned} \lim_{x \to -2^-} g(x) &= \lim_{x \to -2^-} (x + 4) = 2, \\[4pt] \lim_{x \to -2^+} g(x) &= \lim_{x \to -2^+} \left(x^2 - 2\right) = 2. \end{aligned}

\lim_{x \to -2} g(x) = 2 = g(-2)

. All three conditions hold, and

g

is continuous at

-2

x = 2

g(2) = 2^2 - 2 = 2

, and

\begin{aligned} \lim_{x \to 2^-} g(x) &= \lim_{x \to 2^-} \left(x^2 - 2\right) = 2, \\[4pt] \lim_{x \to 2^+} g(x) &= \lim_{x \to 2^+} (7 - x) = 5. \end{aligned}

The one-sided limits differ, so

\lim_{x \to 2} g(x)

doesn't exist, and

g

has a jump discontinuity at

2

The graph of g: no break at x = −2, and a jump at x = 2.

Continuity on an interval

Continuous functions

Polynomials are continuous everywhere. Rational functions, power functions such as

\sqrt{x}

, and trigonometric, exponential, and logarithmic functions are continuous at every point of their domains. At an endpoint of a domain, such as

0

for

\sqrt{x}

, that means continuous from one side.

f

and

g

are continuous at

a

, so are

f + g

f - g

fg

, and, when

g(a) \ne 0

\frac{f}{g}

. If

g

is continuous at

a

and

f

is continuous at

g(a)

, then

f \circ g

is continuous at

a

The rules for sums, differences, products, and quotients also hold with continuity from one side, when

f

and

g

are continuous from the same side. A square root of a continuous function, such as

\sqrt{x - 1}

1

\sqrt{5 - x}

5

, is continuous from the side where it's defined.

Example 3

The graph of

f

from Limits from Graphs and Tables is shown again. Find each

x

(-5, 7)

where

f

is discontinuous, name the first condition of the definition that fails, and classify the discontinuity. Then name the point where giving

f

one new value would make it continuous there, and give that value.

The graph of f on [−5, 7]. The dashed line x = 4 is a vertical asymptote.

x = -2

f(-2) = 3

and

\lim_{x \to -2} f(x) = 1

. Both exist, but they aren't equal, so condition 3 is the first to fail, and the discontinuity is removable.

x = 1

f(1) = 2

and

\lim_{x \to 1^-} f(x) = 2

, but

\lim_{x \to 1^+} f(x) = -1

. The limit doesn't exist, so condition 2 is the first to fail, a jump discontinuity. Since the left-hand limit equals

f(1)

f

is continuous from the left at

1

x = 4

f(4)

isn't defined, so condition 1 fails first. Since

\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = \infty

, the discontinuity is infinite.

At the corner

x = 0

f(0) = 3 = \lim_{x \to 0} f(x)

, so

f

is continuous there. Only the removable discontinuity can be fixed with one value: redefining

f(-2) = 1

makes

f

continuous at

-2

. At

1

and at

4

there is no limit for a new value to match.

Example 4

Find the intervals on which each function is continuous.

\begin{aligned} &\text{(a)}\;\; p(x) = \frac{\sqrt{x - 1}}{x - 4} \\[10pt] &\text{(b)}\;\; q(x) = \ln(x - 2) + \sqrt{5 - x} \\[10pt] &\text{(c)}\;\; r(x) = \frac{1}{1 + \cos x} \end{aligned}

(a) The root

\sqrt{x - 1}

is defined for

x \ge 1

. It's continuous for

x > 1

and from the right at

1

. The denominator is

0

only at

x = 4

. So

p

is continuous on

[1, 4)

and on

(4, \infty)

(b) The logarithm needs

x > 2

, and the root needs

x \le 5

. Each term is continuous where it's defined, the root from the left at

5

, so

q

is continuous on

(2, 5]

(c) The denominator is

0

where

\cos x = -1

, at

x = \pm\pi, \pm 3\pi, \pm 5\pi, \ldots

, the odd multiples of

\pi

. So

r

is continuous on every interval between two consecutive odd multiples of

\pi

, such as

(-\pi, \pi)

and

(\pi, 3\pi)

. Near each odd multiple the denominator approaches

0

through positive values, so each discontinuity is infinite.