Continuity of piece-wise functions

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Here we use limits to ensure piecewise functions are continuous.

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 $x<0$,}\\ \cos (-x) + C &\text {if $x\ge 0$}. \end {cases}$ Find $C$ 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 $\lim _{x\to 0^-} f(x) = \lim _{x\to 0^+}f(x) = f(\answer [given]{0}).$ In this case $\begin{align*} \lim_{x\to 0^-} f(x) &= \lim_{x\to 0^-}\answer[given]{\frac{x}{x-1}}\\ &= \answer[given]{\frac{0}{-1}}\\ &=\answer[given]{0}. \end{align*}$ On there other hand $\begin{align*} \lim_{x\to 0^+} f(x) &= \lim_{x\to 0^+}\left(\answer[given]{\cos(-x)+C}\right)\\ &= \answer[given]{\cos(0) + C}\\ &= \answer[given]{1+ C} \end{align*}$ Hence for our function to be continuous, we need $1 + C = 0\qquad \text {so}\qquad C = \answer [given]{-1}.$ Now, $\displaystyle \lim _{x\to 0} f(x) = f(\answer [given]{0})$ , and so $f$ is continuous.

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 $a$ 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: the definition of continuity will guide you to your goal. To find $a$ and $b$ that make $f$ is continuous at $x=1$ , we need to find $a$ and $b$ such that $\lim _{x\to 1^-} f(x) = \lim _{x\to 1^+}f(x)=f(\answer [given]{1}).$ 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 $\begin{align*} \lim_{x\to 1^+} f(x) &= \lim_{x\to 1^+} \left(\answer[given]{ax^2+bx+2}\right) \\ &= \answer[given]{a+b+2}. \end{align*}$ Hence for this function to be continuous at $x=1$ , we must have that $\begin{align*} 5 &= a+b+2\\ 3 &= \answer[given]{a+b}. \end{align*}$ Hmmmm. At this point, we have 2 unknowns and only one equation. That means we need to find another equation before we’ll be able to solve for both unknowns. Let’s look for that other equation. To find $a$ and $b$ that make $f$ is continuous at $x=3$ , we need to find $a$ and $b$ such that $\lim _{x\to 3^-} f(x) =\lim _{x\to 3^+}f(x) =f(\answer [given]{3}).$ Looking at the limit from the left, we have $\begin{align*} \lim_{x\to 3^-} f(x) &= \lim_{x\to 3^-} \left(\answer[given]{ax^2+bx+2}\right) \\ &=\answer[given]{a\cdot 9 + b\cdot 3 + 2}. \end{align*}$ Looking at the limit from the right, we have $\begin{align*} \lim_{x\to 3^+} f(x) &= \lim_{x\to 3^+} \left(\answer[given]{6x+a-b}\right) \\ &= \answer[given]{18+a-b}. \end{align*}$ Hence for this function to be continuous at $x=3$ , we must have that $\begin{align*} a\cdot 9 + b\cdot 3 + 2 &= 18+a-b\\ a\cdot 8 + b\cdot 4 -16 &= 0\\ a\cdot 2 + b -4 &= 0 \end{align*}$ So now we have two equations and two unknowns: $3=a+b \qquad \text {and}\qquad a\cdot 2 + b -4 = 0.$ Set $b = 3-a$ and write $\begin{align*} 0&= a\cdot 2 + (3-a) -4 \\ &= a -1, \end{align*}$ hence $a=\answer [given]{1}\qquad \text {and so}\qquad b =\answer [given]{2}.$ Let’s check our answers. By 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 $x=3$ .