Continuity

We have seen that any polynomial function P(x) satisfies:

$$\lim_{x \rightarrow a} P(x) = P(a)\;$$

for all real numbers a. This property is known as continuity.

Definition. Let f(x) be a function defined on an interval around a. We say that f(x) is continuous at a iff

$$\lim_{x \rightarrow a} f(x) = f(a)\;.$$

Otherwise, we say that f(x) is discontinuous at a.

Note that the continuity of f(x) at a means two things:

(i)

$$\lim_{x \rightarrow a} f(x)$$ exists,

(ii)

and this limit is f(a).

So to be discontinuous at a, means

(i)

$$\lim_{x \rightarrow a} f(x)$$ does not exist,

(ii)

or if $$\lim_{x \rightarrow a} f(x)$$ exists, then this limit is not equal to f(a).

Basic properties of limits imply the following:

Theorem. If f(x) and g(x) are continuous at a. Then

(1)

f(x) + g(x) is continuous at a;

(2)

$\alpha f(x)$ is continuous at a, where $\alpha$ is an arbitrary number;

(3)

$f(x)\;g(x)$ is continuous at a;

(4)

$$\frac{f(x)}{g(x)}$$ is continuous at a, provided $g(a) \neq 0$;

(5)

If f(x) is positive, i.e. $f(x) \geq 0$, then $\sqrt{f(x)}$ is continuous at a;

(6)

If f(x) is continuous at a and g(x) is continuous at g(a), then their composition $g \circ f(x)$ is continuous at a.

Remark. Many functions are not defined on open intervals. In this case, we can talk about one-sided continuity. Indeed, f(x) is said to be continuous from the left at a iff

$$\lim_{x \rightarrow a-} f(x) = f(a)\;,$$

and f(x) is said to be continuous from the right at a iff

$$\lim_{x \rightarrow a+} f(x) = f(a)\;.$$

Example. The function $f(x) = \sqrt{x}$ is defined for $x \geq 0$. So we can not talk about left-continuity of f(x) at 0. But since

$$\lim_{x \rightarrow 0+} \sqrt{x} = 0\;,$$

we conclude that f(x) is right-continuous at 0.

This concept is also important for step-functions.

Example. Consider the function

$$f(x) = \left\{ \begin{array}{rll} x^3 + 2& \mbox{if $x < 2$}\... ...x = 2$}\\ x^2 + 6 & \mbox{if $x > 2$}\;.\\ \end{array}\right.$$

The details are left to the reader to see

$$\lim_{x \rightarrow 2-} f(x) = \lim_{x \rightarrow 2-} x^3 + 2 = 10,$$

and

$$\lim_{x \rightarrow 2+} f(x) = \lim_{x \rightarrow 2+} x^2 + 6 = 10.$$

So we have

$$\lim_{x \rightarrow 2} f(x) = 10.$$

Since f(2) = 5, then f(x) is not continuous at 2.

Exercise 1. Find A which makes the function

$$f(x) = \left\{ \begin{array}{rll} x^2 - 2& \mbox{if $x < 1$}\\ A x - 4 & \mbox{if $1 \leq x$}\\ \end{array}\right.$$

continuous at x=1.

Answer.

Definition. For a function f(x) defined on a set S, we say that f(x) is continuous on S iff f(x) is continuous for all $a \in S$.

Example. We have seen that polynomial functions are continuous on the entire set of real numbers. The same result holds for the trigonometric functions $\sin(x)$ and $\cos(x)$.

The following two exercises discuss a type of functions hard to visualize. But still one can study their continuity properties.

Exercise 2. Discuss the continuity of

$$f(x) = \left\{ \begin{array}{lll} 1 &\mbox{if $x$ is rational}\\ 0 &\mbox{if $x$ is irrational.}\\ \end{array}\right.$$

Answer.

Exercise 3. Let us modify the previous function: Discuss the continuity of

$$f(x) = \left\{ \begin{array}{lll} \displaystyle \frac{1}{q} &... ...s}\\ &\\ 0 &\mbox{if $x$ is irrational}\\ \end{array}\right.$$

for $x \in (0,1)$. (Two natural numbers p and q are coprime, if their greatest common divisor equals 1.)

Answer.

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