Lesson

Also important in calculus is the idea of continuity. Basically, if we can trace the length of a function without taking our pencil off the paper, the function is continuous.

More formally

A function is continuous at $x=a$`x`=`a` if

(i) $f(a)$`f`(`a`) exists

(ii) $\lim_{x\rightarrow a}f\left(x\right)$lim`x`→`a``f`(`x`) exists and equals $f(a)$`f`(`a`).

There are certain conditions which must exist for a function to be differentiable at $x=a$`x`=`a`

(i) The function is continuous at $x=a$`x`=`a`

(ii) the gradient function is also continuous at $x=a$`x`=`a`.

To say that the gradient function is continuous at $x=a$`x`=`a` is to say that the limit $\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$lim`h`→0`f`(`a`+`h`)−`f`(`a`)`h` exists. (Recall that the 'limit' needs to be the same whether the point $a$`a` is approached from above or from below.)

This limit may also be expressed as $\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$lim`x`→`a``f`(`x`)−`f`(`a`)`x`−`a`.

We see that the notions of continuity and of differentiability are both dependent on the idea of *limit*.

It is usual to give one or more open intervals over which a function is continuous or differentiable. Open intervals are used because it is then possible to have a genuine limit at every point within the interval - where each point may be approached from both sides.

A notation $\lim_{x\rightarrow a^+}$lim`x`→`a`+ is used to indicate an approach to a point $a$`a` from above - a 'right-hand' limit. Similarly, $\lim_{x\rightarrow a^-}$lim`x`→`a`− is used for a 'left-hand' limit.

The function $g$`g` defined on the real numbers and given by $g(x)=x^2$`g`(`x`)=`x`2 is continuous on the open interval notated $(-\infty,\infty)$(−∞,∞). It is also differentiable everywhere within this interval because the gradient function $g'(x)=2x$`g`′(`x`)=2`x` is meaningful for all $x$`x`.

The function $f$`f` defined on the set of real numbers without zero, given by $f(x)=\frac{1}{x}$`f`(`x`)=1`x` is discontinuous as $x=0$`x`=0 because $f(0)$`f`(0) does not exist. As a consequence, $f'(0)$`f`′(0) does not exist. That is, the function is not differentiable at $x=0$`x`=0.

This function is, however, differentiable (and therefore continuous) on the intervals $(-\infty,0)$(−∞,0) and $(0,\infty)$(0,∞).

The rational function $h(x)=\frac{x^3-x^2}{x-1}$`h`(`x`)=`x`3−`x`2`x`−1 has a graph that looks perfectly smooth and continuous. It looks identical to the graph of the function given in Example 1, and yet the function must be undefined at $x=1$`x`=1 since the denominator would then be zero.

If we were interested in the behaviour of this function between the values $x=-10$`x`=−10 and $x=10$`x`=10, we might specify the domain as $\left[-10,1\right)\cup\left(1,10\right]$[−10,1)∪(1,10].

The function is continuous and differentiable on the union of open intervals $\left(-10,1\right)\cup\left(1,10\right)$(−10,1)∪(1,10).

In this case, the discontinuity at $x=1$`x`=1 can be 'removed' by defining the function value at $x=1$`x`=1 to be $1$1. If this is done, the function becomes continuous and differentiable over the whole interval $(-10,10)$(−10,10).

State the interval(s) of the domain over which the function is continuous.

Loading Graph...

$($( $\editable{}$, $\editable{}$ ]

Find $\lim_{x\to7^-}\left(\frac{\left|x-7\right|}{x-7}\right)$lim`x`→7−(|`x`−7|`x`−7).

Consider the function $f\left(x\right)=\frac{x}{\left(1-x\right)^3}$`f`(`x`)=`x`(1−`x`)3.

Find $\lim_{x\to1^+}f\left(x\right)$lim

`x`→1+`f`(`x`).Find $\lim_{x\to1^-}f\left(x\right)$lim

`x`→1−`f`(`x`).

Identify discontinuities and limits of functions

Apply differentiation methods in solving problems