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. Recall that a function $f$ is continuous at a point $a$ if $\displaystyle \lim _{x\to a}f(x) = f(a)$. This gave the continuity checklist. To determine whether a function $f$ is continuous at a point $a$, we must check that:

(a): $f(a)$ is defined,
(b): $\lim _{x\to a} f(x)$ exists, and then
(c): $\lim _{x\to a} f(x) = f(a)$.

In this section, since we are dealing with piecewise-defined functions, we may have to consider one-sided limits when determining if item cc:lim holds for our function. However, item cc:lim does not mention one-sided limits. When using the checklist to justify the continuity of a function, take care to mention the specific conditions that are required.

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 $x=0$.

The question asks us to find the value of $c$ so the function is continuous at $x=0$, so we use the continuity checklist. We start with item cc:val. Since $f(0) = \answer [given]{0}$ exists, this item is satisfied.

For item cc:lim, 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 $x>0$).

\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 $\displaystyle \lim _{x\to 0} f(x)$ to exist,

\[ 1 + c = 0\qquad \text {so}\qquad c = \answer [given]{-1}. \]

So, if $c=-1$, the limit $\displaystyle \lim _{x\to 0} f(x)=0$ which means item cc:lim is satisfied.

For item cc:conc we need to check that the value we found for $f(0)$ is the same as the value we found for $\displaystyle \lim _{x\to 0}f(x)$. These values agree, so item cc:conc is satisfied. This proves that $f$ is continuous at $x=0$.

In the previous example, notice that finding the value of $c$ is not where the work ended. In order to justify the answer, we have to verify that all three conditions of the checklist are satisfied for the value we found.

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

\[ \lim _{x\to 1} f(x) =f(\answer [given]{1}). \]

Let us begin with item cc:val of the continuity checklist at $x=1$. $f(1)=a+b+2$ exists for all values of $a$ and $b$, so this item is satisfied.

We now check item cc:lim of the checklist by verifying that $\displaystyle \lim _{x\to 1}f(x)$ exists. In order to calculate $\lim _{x\to 1}f(x)$, 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*}

Item cc:conc is then satisfied at the point $x=1$ whenever $a+b=3$.

To find $a$ and $b$ that make $f$ is continuous at $x=3$, we need to find $a$ and $b$ such that

\[ \lim _{x\to 3} f(x) =f(3). \]

Let us return to the continuity checklist at the point $x=3$. For item cc:val notice that $f(3)=18+a-b$, which exists for all values of $a$ and $b$.

To verify item cc:lim we must check that $\lim _{x\to 3}f(x)$ exists. Again, the formula for $f(x)$ changes around $x=3$ so we check the one-sided limits. From the left:

\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*}

In order for $\lim _{x\to 3}f(x)$ to exist, these must be equal.

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

This means item cc:lim is satisfied for $a$ and $b$ with $2a+b=4$.

Item cc:conc is then satisfied for all $a$ and $b$ with $2a+b=4$, since this gives $f(3) = 18+a+b$ and $\lim _{x\to 3}f(x)=18+a+b = 9a+3b+2$.

So in order to be continuous at BOTH $x=1$ AND $x=3$, 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

\[ \lim _{x\to 1} f(x) = f(1) = 5, \]

and

\[ \lim _{x\to 3} f(x) = f(3) = 17. \]

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 $y=f(x)$:

\[ \graph [xmin=-10,xmax=10,ymin=0,ymax=30]{y=x+4\left \{x<1\right \},y=x^2+2x+2\left \{1\leq x<3\right \},y=6x-1\left \{3 \leq x\right \}} \]