Continuity

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Continuity is defined by limits.

A few sections ago, we discussed what we called ’nice’ functions. These were functions that, at a particular $x$ -value $a$ , had easy-to-compute limits. In fact, computing the limit of a ’nice’ function is as simple as plugging $a$ into the function itself so $\lim _{x\to a} f(x) = f(a).$ In other words, if a function is what we called ’nice’ at $x=a$ , the limit of that function at $x=a$ is equal to the function value at $x=a$ . As you might imagine, this property is pretty useful, so we give it a name: continuity.

A function $f$ is continuous at a point $a$ if $\lim _{x\to a}f(x) = f(a).$

Consider the graph of $y=f(x)$ below

Which of the following are true? (Select all that are true)

$f$ is continuous at $x=0.5$ $f$ is continuous at $x=1$ $f$ is continuous at $x=1.5$

It is very important to note that saying

‘‘a function $f$ is continuous at a point $a$ ’’

is really making three statements:

(a): $f(a)$ is defined. That is, $a$ is in the domain of $f$ .
(b): $\displaystyle \lim _{x\to a} f(x)$ exists.
(c): $\displaystyle \lim _{x\to a} f(x) = f(a)$ .

The first two of these statements are implied by the third statement.

Find the discontinuities (the values of $x$ where the function is not continuous) for the function described below:

To start, $f$ is not even defined at $x=\answer [given]{2}$ , hence $f$ cannot be continuous at $x=\answer [given]{2}$ .

Next, from the plot above we see that $\displaystyle \lim _{x\to 4} f(x)$ does not exist because $\lim _{x\to 4^-}f(x) = \answer [given]{1}\qquad \text {and}\qquad \lim _{x\to 4^+}f(x) \approx \answer [given,tolerance=1]{3.5}$ Hence $\lim _{x\to 4} f(x) \ne f(4)$ , and so $f(x)$ is not continuous at $x=4$ .

We also see that $\displaystyle \lim _{x\to 6} f(x) \approx \answer [given,tolerance=.5]{3}$ while $f(6) = \answer [given]{2}$ . Hence $\displaystyle \lim _{x\to 6} f(x)$ does not exist, and so $f$ is not continuous at $x=6$ .

Building from the definition of continuity at a point, we can now define what it means for a function to be continuous on an open interval.

A function $f$ is continuous on an open interval $I$ if $\displaystyle \lim _{x\to a} f(x) = f(a)$ for all $a$ in $I$ .

Loosely speaking, a function is continuous on an interval $I$ if you can draw the function on that interval without any breaks in the graph. This is often referred to as being able to draw the graph ‘‘without picking up your pencil.’’

Continuity of Common Functions The following functions are continuous on the given intervals for $k$ a real number and $b$ a positive real number:

Constant function: $f(x) =k$ is continuous on $-\infty < x < \infty$ .
Identity function: $f(x) = x$ is continuous on $-\infty < x < \infty$ .
Power function: $f(x)=x^b$ is continuous on $-\infty < x < \infty$ .
Sine and cosine: Both $\sin (x)$ and $\cos (x)$ are continuous on $-\infty < x < \infty$ .

In essence, we are saying that the functions listed above are continuous wherever they are defined, that is, on their natural domains.

Compute: $\displaystyle \lim _{x\to 3} x^\pi \begin {prompt}= \answer {3^\pi }\end {prompt}$

The function $f(x)=x^\pi$ is of the form $x^k$ for a real number $k$ . Therefore, $f(x)=x^\pi$ is continuous for all real values of $x$ . In particular, $f(x)$ is continuous at $x=3$ . Since $x^\pi$ is continuous at $3$ , we know that $\displaystyle \lim _{x\to 3} f(x) = f(3)$ . That is, $\displaystyle \lim _{x\to 3} x^\pi = 3^\pi$

True or false: If $f$ and $g$ are continuous functions on an interval $I$ , then $f\pm g$ is continuous on $I$ .

True False

This follows from the Sum/Difference Law.

True or false: If $f$ and $g$ are continuous functions on an interval $I$ , then $f/g$ is continuous on $I$ .

True False

In this case, $f/g$ will not be continuous for $x$ where $g(x) = 0$ .

Continuity of Polynomial Functions All polynomial functions, meaning functions of the form $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ where $n$ is a positive whole number and each $a_i$ is a real number, are continuous for all real numbers.

Continuity of Rational Functions Let $f$ and $g$ be polynomials. Then a rational function, meaning an expression of the form $h=\frac {f}{g}$ is continuous for all real numbers except where $g(x)=0$ . That is, rational functions are continuous wherever they are defined.

Let $a$ be a real number such that $g(a)\neq 0$ . Then, since $g(x)$ is continuous at $a$ , $\displaystyle \lim _{x\to a} g(x) \neq 0$ . Therefore, write with me, $\displaystyle \lim _{x \to a} h(x) = \displaystyle \lim _{x\to a} \frac {f(x)}{g(x)}$ and now by the Quotient Law, $\frac {\displaystyle \lim _{x\to a} f(x)}{\displaystyle \lim _{x\to a} g(x)}$ and by the continuity of polynomials we may now set $x=a$ $\frac {f(a)}{g(a)}=h(a).$ Since we have shown that $\displaystyle \lim _{x\to a} h(x) = h(a)$ , we have shown that $h$ is continuous at $x=a$ .

Where is $f(x) = \frac {x^2-3x+2}{x-2}$ continuous?

for all real numbers at $x=2$ for all real numbers, except $x=2$ impossible to say

Now, we give basic rules for how limits interact with composition of functions. Before continuing on, you may wish to review what is meant by a composition of functions.

Composition Limit Law If $f(x)$ is continuous at $x = \displaystyle \lim _{x\to a} g(x)$ , then $\displaystyle \lim _{x\to a} f(g(x)) = f(\displaystyle \lim _{x\to a} g(x)).$

Because the limit of a continuous function is the same as the function value, we can now pass limits inside continuous functions.

Continuity of Composite Functions If $g$ is continuous at $x=a$ , then $f(g(x))$ is continuous at $x=a$ .

Compute the following limit using limit laws: $\displaystyle \lim _{x \to 0} \sqrt {\cos (x)}$

By continuity of $x^k$ , assuming $\displaystyle \lim _{x \to 0} \cos (x) >0$ , $\displaystyle \lim _{x \to 0} \sqrt {\cos (x)} = \sqrt {\displaystyle \lim _{x \to 0} \cos (x)},$ and now since cosine is continuous for all real numbers, $\sqrt {\cos (0)} = \sqrt {1} = 1.$

Left and right continuity

At this point we have a small problem. For functions such as $\sqrt {x}$ , the natural domain is $0\leq x <\infty$ . This is not an open interval. What does it mean to say that $\sqrt {x}$ is continuous at $0$ when $\sqrt {x}$ is not defined for $x<0$ ? To get us out of this quagmire, we need a new definition:

A function $f$ is left continuous at a point $a$ if $\displaystyle \lim _{x\to a^-} f(x) = f(a)$ .

A function $f$ is right continuous at a point $a$ if $\displaystyle \lim _{x\to a^+} f(x) = f(a)$ .

Now we can say that a function is continuous at a left endpoint of an interval if it is right continuous there, and a function is continuous at the right endpoint of an interval if it is left continuous there. This allows us to talk about continuity on closed intervals.

Here we give the graph of a function defined on $[0,10]$ .

What are the largest intervals of continuity for this function?

$[0,10]$ $[0,4)$ and $(4,10]$ $[0,4]$ , $[4,6]$ , and $[6,10]$ $(0,4)$ , $(4,6)$ , and $(6,10)$ $[0,4]$ , $(4,6)$ , and $[6,10]$ $[0,4]$ , $(4,6)$ , and $(6,10]$ $[0,4)$ , $(4,6)$ , and $(6,10]$ $(0,4]$ , $[4,6]$ , and $[6,10)$

Notice that our function is left continuous at $x=4$ so we can include $4$ in the interval $[0,4]$ . Four is not included in the interval $(4,6)$ because our function is not right continuous at $x=4$ . Similarly, our function is neither right or left continuous at $x=6$ , so $6$ is not included in any intervals. Our function is left continuous at $x=0$ and right continuous at $x=10$ so we included these endpoints in our intervals.