Continuity Theorems, calculus lesson

Checking continuity straight from the definition every time would be exhausting. Fortunately continuity survives the usual ways of building functions. Once you know a few basic functions are continuous, everything you can build from them is too, and that is why plugging in the point works for so many limits.

Key terms

Algebra of continuous functions

Definition Suppose $f$ and $g$ are both continuous at $c$ and $k$ is a constant. Then the following functions are continuous at $c$ .

f + g, \qquad f - g, \qquad k f, \qquad f g, \qquad \frac{f}{g} \ \text{ provided } g(c) \neq 0.

The proof is one line for each, using the limit laws. For the product it reads

\lim_{x \to c} (fg)(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) = f(c) g(c).

The same rules hold for one-sided continuity.

In plain words Add, subtract, multiply or divide continuous functions and the result is still continuous, as long as you never divide by zero.

Example The functions $x$ and constants are continuous everywhere, so by repeated products and sums every polynomial is continuous everywhere. So the rational function

R(x) = \frac{x^3 + 1}{x^2 - 4}

is continuous at every $x$ except $x = 2$ and $x = -2$ , where the denominator is zero. In particular

\lim_{x \to 1} R(x) = R(1) = \frac{2}{-3} = -\frac{2}{3}.

Compositions

Definition If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then $f \circ g$ , given by $(f \circ g)(x) = f(g(x))$ , is continuous at $c$ . More generally, suppose $\lim_{x \to c} g(x) = b$ , $f$ is continuous at $b$ , and $f(g(x))$ is defined for all $x \neq c$ in some open interval around $c$ . Then

\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x)\right) = f(b).

In plain words A continuous function can be moved past a limit sign. Feed a continuous inner function into a continuous outer function and the chain stays unbroken.

Example With $g(x) = x^2 + 9$ and $f(u) = \sqrt{u}$ , both continuous, the limit is

\lim_{x \to 4} \sqrt{x^2 + 9} = \sqrt{25} = 5.

The second form also handles inner functions with a hole. The inner limit is

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2,

the quotient is positive for $x$ near $1$ , and $\sqrt{u}$ is continuous at $2$ , so

\lim_{x \to 1} \sqrt{\frac{x^2 - 1}{x - 1}} = \sqrt{2}.

Continuity of elementary functions

Definition The following functions are continuous at every point of their domains. These are polynomials, rational functions, root functions $\sqrt[n]{x}$ , the trigonometric functions $\sin x$ , $\cos x$ , $\tan x$ and the others, the inverse trigonometric functions, exponential functions $a^x$ with $a > 0$ , and logarithms $\log_a x$ with $a > 0$ , $a \neq 1$ . Every function built from these by finitely many sums, differences, products, quotients and compositions is continuous at every point of its domain. At an endpoint of the domain this means continuity from the inside.

In plain words For any formula made of the usual functions, you can find the limit at a point of the domain by plugging the point in.

Example The function $h(x) = \ln(x^2 - 1)$ is defined when $x^2 - 1 > 0$ , that is on $(-\infty, -1) \cup (1, \infty)$ , and it is continuous there. So $\lim_{x \to 2} \ln(x^2 - 1) = \ln 3$ . Similarly, since the denominator below is never zero,

\lim_{x \to \pi} \frac{x \cos x}{2 + \sin x} = \frac{\pi \cdot (-1)}{2 + 0} = -\frac{\pi}{2}.

Continuous images of intervals

Definition If $f$ is continuous on an interval $I$ , then its image $f(I) = \{f(x) \mid x \in I\}$ is also an interval, possibly a single point. If $I = [a, b]$ is closed and bounded, then $f(I)$ is a closed bounded interval $[m, M]$ . For intervals of other kinds the image can be open, closed or half-open.

In plain words A continuous function cannot tear an interval apart. The outputs fill a whole interval with no gaps, and a closed bounded interval goes to a closed bounded interval.

Example For $f(x) = x^2$ on $[-1, 2]$ the image is $[0, 4]$ , with $0 = f(0)$ and $4 = f(2)$ . On the open interval $(-1, 2)$ the image is $[0, 4)$ . It still contains $0$ , but $4$ is now missing because $x = 2$ is no longer allowed. So an open interval does not have to go to an open interval.

The max-min theorem

Definition If $f$ is continuous on a closed bounded interval $[a, b]$ , then $f$ attains an absolute maximum value $M$ and an absolute minimum value $m$ on $[a, b]$ . That is, there are numbers $x_1$ and $x_2$ in $[a, b]$ with

f(x_1) = m \le f(x) \le M = f(x_2) \quad \text{for every } x \in [a, b].

Each hypothesis matters. The interval must be closed, it must be bounded, and $f$ must be continuous on all of it.

In plain words On a closed bounded interval a continuous function has a highest point and a lowest point that it actually reaches. The theorem says they exist, not where they are.

Example Take $f(x) = x^2 - 2x + 3$ on $[0, 3]$ . Completing the square gives $f(x) = (x - 1)^2 + 2$ , so the minimum is $2$ at $x = 1$ and the maximum is $f(3) = 6$ . Now drop one hypothesis at a time. On the open interval $(0, 1)$ the function $x$ has no largest or smallest value. On the unbounded interval $[0, \infty)$ the function $x$ has no maximum. On $[0, 1]$ the function equal to $x$ for $x < 1$ and to $0$ at $x = 1$ is not continuous at $1$ , and it has no maximum, because its values come as close to $1$ as you like but never reach it. In each case the conclusion fails.

Check yourself

Where is $f(x) = \frac{\sqrt{x}}{x - 4}$ continuous?

Show the answer

The root $\sqrt{x}$ needs $x \ge 0$ and the denominator needs $x \neq 4$ . So $f$ is continuous on $[0, 4) \cup (4, \infty)$ , from the right at $0$ .

Compute $\lim_{x \to 3} \sqrt{x^2 + 7}$ and say which theorem justifies plugging in.

Show the answer

The limit is $\sqrt{16} = 4$ . The inner polynomial is continuous at $3$ and the square root is continuous at $16$ , so the composition is continuous at $3$ .

Does $f(x) = \frac{1}{x}$ have a largest value among the inputs $x$ in $[-1, 1]$ with $x \neq 0$ ? Why does the max-min theorem not apply?

Show the answer

No, its values grow without bound as $x \to 0^+$ . The theorem needs $f$ continuous at every point of $[-1, 1]$ , and $f$ is not even defined at $0$ .

Continuity Theorems

Theory, part 33

Practice, part 34

Key terms

Algebra of continuous functions

Compositions

Continuity of elementary functions

Continuous images of intervals

The max-min theorem

Check yourself

How this lesson connects

Builds on