Continuity

Functions: Introduction to functions

Continuity

Let $f$ be a real function defined on an open interval around a point $a$ .

If $\displaystyle \lim_{x\to a}f(x) =f(a)$ , then $f$ is called continuous at $a$ . If not, then $f$ is called discontinuous at $a$ .

If $f$ is continuous at every point of an open interval $\ivoo{c}{d}$ , then $f$ is called continuous on $\ivoo{c}{d}$ .

In other words, a function is continuous if the graph can be drawn without lifting the pen from the paper.

If $a$ is a point of $\ivoo{c}{d}$ such that $\displaystyle \lim_{x\uparrow a}f(x) =f(a)$ , then $f$ is called left continuous at $a$ .

If $a$ is a point of $\ivoo{c}{d}$ such that $\displaystyle \lim_{x\downarrow a}f(x) =f(a)$ , then $f$ is called right continuous at $a$ .

A point must meet both conditions in order to establish that $f$ is continuous at $a$ .

We show a function that is continuous everywhere except at the point $0$ . This is due to a jump in the graph of $f$ . The function is the Heaviside function $H$ defined by $H(x) = 0$ if $x\lt0$ and $H(x) = 1$ if $x\gt0$ . The graph of this function shows a jump at $0$ and so is discontinuous there.

Consider the function $f$ given by $\displaystyle f(x)=\begin{cases} 8 x+9&\text{if }x\lt1\\ 9 x^2+8&\text{if }x\geq1\end{cases}$

Is this function continuous at $1$ ?

Yes

The function $f$ is left continuous at $x=1$ :
$\begin{array}{rcl} \displaystyle\lim_{x\uparrow1}f(x)&=&\lim_{x\uparrow1}( 8x+9)=8+9=17=f(1)\end{array}$
The function $f$ is right continuous at $x=1$ :
$\begin{array}{rcl}\displaystyle\lim_{x\downarrow1}f(x)&=&\lim_{x\downarrow1} (9 x^2+8)=9+8=17=f(1)\end{array}$
Since $f$ is both right continuous and left continuous at $1$ , it is continuous at $1$ .

Below the graph of $f$ is drawn. It has no jump at $1$ .

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