Continuity of piecewise functions

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Here we use limits to check whether piecewise functions are continuous.

Before we start talking about continuity of piecewise functions, let’s remind ourselves of all famous functions that are continuous on their domains.

Continuity of Famous Functions The following functions are continuous on their natural domains:

Constant functions
Polynomials
Rational functions
Power functions
Exponential functions
Logarithmic functions
Trigonometric functions
Inverse trigonometric functions

In essence, we are saying that the functions listed above are continuous wherever they are defined.

We proved continuity of polynomials earlier using the Sum Law, Product Law and continuity of power functions.

We proved continuity of rational functions earlier using the Quotient Law and continuity of polynomials.

We can prove continuity of the remaining four trig functions using the Quotient Law and continuity of sine and cosine functions.

Since a continuous function and its inverse have “unbroken” graphs, it follows that an inverse of a continuous function is continuous on its domain.

This implies that inverse trig functions are continuous on their domains.

Using the Limit Laws we can prove that given two functions, both continuous on the same interval, then their sum, difference, product, and quotient (where defined) are also continuous on the same interval (where defined).

In this section we will work a couple of examples involving limits, continuity and piecewise functions.

Consider the following piecewise defined function $f(x) = \begin {cases} \frac {x}{x-1} &\text {if $x<0$,}\\ e^{-x} + c &\text {if $x\ge 0$}. \end {cases}$ Find the constant $c$ so that $f$ is continuous at $x=0$ .

To find $c$ such that $f$ is continuous at $x=0$ , we need to find $c$ such that $\lim _{x\to 0} f(x) = f\Bigl (\answer [given]{0}\Bigr ).$ In this case, in order to compute the limit, we will have to compute two one-sided limits, since the expression for $f(x)$ if $x<0$ is different from the expression for $f(x)$ if $x>0$ . So, $\begin{align*} \lim_{x\to 0^-} f(x) &= \lim_{x\to 0^-}\answer[given]{\frac{x}{x-1}}\\ &= \frac{\answer[given]{0}}{-1}\\ &=\answer[given]{0}. \end{align*}$

and

$\begin{align*} \lim_{x\to 0^+} f(x) &= \lim_{x\to 0^+}\left(e^{-x}+c\right)\\ &= e^{\answer[given]{0}} + c\\ &= \answer[given]{1+ c} \end{align*}$

In order for $f$ to be continuous, the limit $\lim _{x\to 0} f(x)$ has to exist. This means that $1 + c = 0\qquad \text {so}\qquad c = \answer [given]{-1}.$ So, if $c=-1$ , the limit $\lim _{x\to 0} f(x)=0$ . Now, we have find the value

$f(0) = \answer [given]{0}.$ Therefore, $\lim _{x\to 0} f(x) = f\Bigl (0\Bigr )\checkmark ,$ which proves that $f$ is continuous at $x=0$ .

Consider the next, more challenging example.

Consider the following piecewise defined function $f(x) = \begin {cases} x+4 &\text {if $x<1$,}\\ ax^2+bx+2 &\text {if $1\le x< 3$,}\\ 6x+a-b &\text {if $x\ge 3$.} \end {cases}$ Find the constants $a$ and $b$ so that $f$ is continuous at both $x=1$ and $x=3$ .

This problem is more challenging because we have more unknowns. However, be brave intrepid mathematician. To find $a$ and $b$ that make $f$ continuous at $x=1$ , we need to find $a$ and $b$ such that $\lim _{x\to 1} f(x) =f(\answer [given]{1}).$ Since $f(1)=a+b+2$ , it follows that $\lim _{x\to 1} f(x) =a+b+2.$ We have to compute two one-sided limits, since the expression for $f(x)$ if $x<1$ is different from the expression for $f(x)$ if $x>1$ . Looking at the limit from the left, we have $\begin{align*} \lim_{x\to 1^-} f(x) &= \lim_{x\to 1^-} \left(\answer[given]{x+4}\right) \\ &=\answer[given]{5}. \end{align*}$

Looking at the limit from the right, we have

$\begin{align*} \lim_{x\to 1^+} f(x) &= \lim_{x\to 1^+} \left(ax^2+bx+2\right) \\ &= \answer[given]{a+b+2}. \end{align*}$

Hence, for the limit $\lim _{x\to 1} f(x)$ to exist, we must have that

$\begin{align*} 5 &= a+b+2\\ or\\ a+b &= \answer[given]{3}. \end{align*}$

Also, Hmmmm. More work needs to be done.

To find $a$ and $b$ that make $f$ is continuous at $x=3$ , we need to find $a$ and $b$ such that $\lim _{x\to 3} f(x) =f(3).$ Since $f(3)=18+a-b$ , it follows that $\lim _{x\to 3} f(x) =18+a-b.$ Looking at the limit from the left, we have

$\begin{align*} \lim_{x\to 3^-} f(x) &= \lim_{x\to 3^-} \left(ax^2+bx+2\right) \\ &=a\cdot 9 + b\cdot 3 + 2. \end{align*}$

Looking at the limit from the right, we have

$\begin{align*} \lim_{x\to 3^+} f(x) &= \lim_{x\to 3^+} \left(6x+a-b\right) \\ &= \answer[given]{18+a-b}. \end{align*}$

Hence

$\begin{align*} 9a + 3b + 2 &= 18+a-b\\ 8a + 4b -16 &= 0\\ 2a + b -4 &= 0 \end{align*}$

So now we have two equations and two unknowns: $a+b=3 \qquad \text {and}\qquad 2a + b = 4.$ Set $b = 3-a$ and write

$\begin{align*} 2a+3-a&=4 \\ a &= 1, \end{align*}$

hence $a=\answer [given]{1}\qquad \text {and so}\qquad b =\answer [given]{2}.$ Let’s check, so now plugging in values for both $a$ and $b$ we find $f(x) = \begin {cases} x+4 &\text {if $x<1$,}\\ \answer [given]{x^2+2x+2} &\text {if $1\le x< 3$,}\\ \answer [given]{6x-1} &\text {if $x\ge 3$.} \end {cases}$ Now $\lim _{x\to 1^-} f(x) =\lim _{x\to 1^+}f(x) =f(1) = 5,$ and $\lim _{x\to 3^-} f(x) =\lim _{x\to 3^+}f(x) =f(3) = 17.$ So setting $a= \answer [given]{1}$ and $b=\answer [given]{2}$ makes $f$ continuous at $x=1$ and $x=3$ . We can confirm our results by looking at the graph of $y=f(x)$ : $\graph [xmin=-10,xmax=10,ymin=0,ymax=30]{y=x+4\left \{x<1\right \},y=x^2+2x+2\left \{1\leq x<3\right \},y=6x-1\left \{3 \leq x\right \}}$