### Functions: Introduction to functions

### Continuity

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.

Is this function continuous at #1#?

The function #f# is left continuous at #x=1# :

\[

\begin{array}{rcl}

\displaystyle\lim_{x\uparrow1}f(x)&=&\lim_{x\uparrow1}( 4x+3)=4+3=7=f(1)\end{array}\]

The function #f# is right continuous at #x=1#:

\[\begin{array}{rcl}\displaystyle\lim_{x\downarrow1}f(x)&=&\lim_{x\downarrow1} (3 x^2+4)=3+4=7=f(1)\end{array}\]

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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