14.03 Limits and continuity

Also important when working with limits 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. However, we would like to be able to prove a function is continuous at a point mathematically.

We see that the notion of continuity is dependent on the idea of limit.

It is common to give one or more open intervals over which a function is continuous. 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 $\lim_{x\rightarrow a^+}$limx→a+ is used to indicate an approach to a point $a$a from above - a 'right-hand' limit. Similarly, $\lim_{x\rightarrow a^-}$limx→a− is used for a 'left-hand' limit.

For a limit to exist, $\lim_{x\rightarrow a^+}f\left(x\right)=\lim_{x\rightarrow a^-}f\left(x\right)$limx→a+f(x)=limx→a−f(x).

Exploration

The rational function $h(x)=\frac{x^3-x^2}{x-1}$h(x)=x3−x2x−1​ has a graph that looks perfectly smooth and continuous. It looks almost identical to the graph of the function $f(x)=x^2$f(x)=x2, since $h(x)=\frac{x^3-x^2}{x-1}=\frac{x^2(x-1)}{x-1}$h(x)=x3−x2x−1​=x2(x−1)x−1​, and yet the function will be undefined at $x=1$x=1 since the denominator would then be zero. 

If we were interested in the behavior 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 on the union of open intervals $\left(-10,1\right)\cup\left(1,10\right)$(−10,1)∪(1,10). 

If we wanted to have $h(x)$h(x) be continuous, we could define $h(1)$h(1) using a piecewise function.

$h\left(x\right)$h(x)	$=$=		$\frac{x^3-x^2}{x-1}$x3−x2x−1​	when $x\ne1$x≠1
			$1$1	when $x=1$x=1

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

Worked example

Question 1

For the graph below prove that the graph is not continuous.

Think: We need to show that at least one of the three requirements above is not true. Let's start with the first requirement and work our way through. We can see that the point we need to focus on is $x=0$x=0.

Do:

1. Does $f(0)$f(0) exist?
   Yes, from the graph $f(0)=1$f(0)=1 as it is where there is a closed endpoint.
    
2. Does $\lim_{x\rightarrow a}f\left(x\right)$limx→af(x) exist? In particular, we need to show that $\lim_{x\rightarrow0^+}f\left(x\right)=\lim_{x\rightarrow0^-}f\left(x\right)$limx→0+f(x)=limx→0−f(x). From the graph, we can read off the right and left hand limits.

Left hand limit: $\lim_{x\rightarrow0^-}f\left(x\right)=-1$limx→0−f(x)=−1

Right hand limit: $\lim_{x\rightarrow0^+}f\left(x\right)=1$limx→0+f(x)=1

 $\lim_{x\rightarrow0^+}f\left(x\right)\ne\lim_{x\rightarrow0^-}f\left(x\right)$limx→0+f(x)≠limx→0−f(x), so $\lim_{x\rightarrow0}f\left(x\right)$limx→0f(x) does not exist.

Since the second requirement for continuity does not hold, we can say that $f(x)$f(x) is not continuous.

Reflect: Is there any way we could "remove" this discontinuity or is it a different type of discontinuity?

Practice questions

Question 2

Question 3

question 4