Continuous and discontinuous functions

A function $f(x)$ is said to be continuous for $x = a$ if the limiting value of the function when $x$ approaches the limit $a$ in any manner is the value assigned to the function for $x = a$. In symbols, if

$$\lim_{x \to a} f(x) = f(a),$$

then $f(x)$ is continuous for $x = a$.

The function is said to be discontinuous for $x = a$ if this condition is not satisfied. For example, if

$$\lim_{x \to a} f(x) = \infty,$$

the function is discontinuous for $x = a$.

The attention of the student is now called to the following cases which occur frequently.

CASE I. As an example illustrating a simple case of a function continuous for a particular value of the variable, consider the function

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

For $x=1$, $f(x) = f(l) = 3$. Moreover, if $x$ approaches the limit $1$ in any manner, the function $f(x)$ approaches $3$ as a limit. Hence the function is continuous for $x=1$.

CASE II. The definition of a continuous function assumes that the function is already defined for $x = a$. If this is not the case, however, it is sometimes possible to assign such a value to the function for $x = a$ that the condition of continuity shall be satisfied. The following theorem covers these cases.

Theorem 3.6.1   If $f(x)$ is not defined for $x = a$, and if

$$\lim_{x \to a} f(x) = B,$$

then $f(x)$ will be continuous for $x = a$, if $B$ is assumed as the value of $f(x)$ for $x = a$.

Thus the function

$$\frac{x^2 - 4}{x - 2}$$

is not defined for $x = 2$ (since then there would be division by zero). But for every other value of $x$,

$$\frac{x^2 - 4}{x + 2} = x + 2;$$

and

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

therefore $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$. Although the function is not defined for $x = 2$, if we assign it the value $4$ for $x = 2$, it then becomes continuous for this value.

A function $f(x)$ is said to be continuous in an interval when it is continuous for all values of $x$ in this interval^3.3.