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

\begin{displaymath}\lim_{x \rightarrow a} P(x) = P(a)\;\end{displaymath}

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

\begin{displaymath}\lim_{x \rightarrow a} f(x) = f(a)\;.\end{displaymath}

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

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

$\displaystyle \lim_{x \rightarrow a} f(x)$ exists,
and this limit is f(a).
So to be discontinuous at a, means
$\displaystyle \lim_{x \rightarrow a} f(x)$ does not exist,
or if $\displaystyle \lim_{x \rightarrow a} f(x)$ 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

f(x) + g(x) is continuous at a;
$\alpha f(x)$ is continuous at a, where $\alpha$ is an arbitrary number;
$f(x)\;g(x)$ is continuous at a;
$\displaystyle \frac{f(x)}{g(x)}$ is continuous at a, provided $g(a) \neq 0$;
If f(x) is positive, i.e. $f(x) \geq 0$, then $\sqrt{f(x)}$ is continuous at a;
If f(x) is continuous at a and g(x) is continuous at f(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

\begin{displaymath}\lim_{x \rightarrow a-} f(x) = f(a)\;,\end{displaymath}

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

\begin{displaymath}\lim_{x \rightarrow a+} f(x) = f(a)\;.\end{displaymath}

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

\begin{displaymath}\lim_{x \rightarrow 0+} \sqrt{x} = 0\;,\end{displaymath}

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

This concept is also important for step-functions.

Example. Consider the function

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

The details are left to the reader to see

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


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

So we have

\begin{displaymath}\lim_{x \rightarrow 2} f(x) = 10.\end{displaymath}

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

Exercise 1. Find A which makes the function

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

continuous at x=1.


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

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


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

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

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


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