Piecewise functions are functions that use different rules on different intervals. When finding limits and checking continuity, the key idea is to look at the rule that applies from the left side and the rule that applies from the right side. These problems include one-sided limits, two-sided limits, function values, continuity checks, and solving for constants that make a piecewise function continuous.
Practice Problems
\(\textbf{1)}\) Find \( \displaystyle \lim_{x\to 2^{-}} f(x) \)
where \(f(x) = \begin{cases}
5x+3 & \text{if } x \lt 2 \\
4x & \text{if } x \geq 2
\end{cases}\)
The limit is \( 13 \)
\(\,\,\,\,\,\text{Since }x\to2^{-}\text{, use the rule for }x<2.[/latex]
[latex]\,\,\,\,\,f(x)=5x+3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{-}}f(x)=5(2)+3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{-}}f(x)=13\)
\(\,\,\,\,\,\)The limit is \(13\)
\(\textbf{2)}\) Find \( \displaystyle \lim_{x\to 2^{+}} f(x) \)
where \(f(x) = \begin{cases}
5x+3 & \text{if } x \lt 2 \\
4x & \text{if } x \geq 2
\end{cases}\)
The limit is \( 8 \)
\(\,\,\,\,\,\text{Since }x\to2^{+}\text{, use the rule for }x\geq2.\)
\(\,\,\,\,\,f(x)=4x\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{+}}f(x)=4(2)\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{+}}f(x)=8\)
\(\,\,\,\,\,\)The limit is \(8\)
\(\textbf{3)}\) Find \( \displaystyle \lim_{x\to 2} f(x) \)
where \(f(x) = \begin{cases}
5x+3 & \text{if } x \lt 2 \\
4x & \text{if } x \geq 2
\end{cases}\)
The limit does not exist
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{-}}f(x)=5(2)+3=13\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2^{+}}f(x)=4(2)=8\)
\(\,\,\,\,\,13\neq8\)
\(\,\,\,\,\,\text{Since the left and right limits are not equal, the two-sided limit does not exist.}\)
\(\,\,\,\,\,\)The limit does not exist
\(\textbf{4)}\) Find \( f(2) \)
where \(f(x) = \begin{cases}
5x+3 & \text{if } x \lt 2 \\
4x & \text{if } x \geq 2
\end{cases}\)
The answer is \(8\)
\(\,\,\,\,\,\text{Since }x=2\text{, use the rule that includes }x=2.\)
\(\,\,\,\,\,f(x)=4x\text{ if }x\geq2\)
\(\,\,\,\,\,f(2)=4(2)\)
\(\,\,\,\,\,f(2)=8\)
\(\,\,\,\,\,\)The answer is \(8\)
\(\textbf{5)}\) Find \( \displaystyle \lim_{x\to 4^{-}} f(x) \)
where \(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
The limit is \( 1 \)
\(\,\,\,\,\,\text{Since }x\to4^{-}\text{, use the rule for }x\leq4.\)
\(\,\,\,\,\,f(x)=-x+5\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{-}}f(x)=-(4)+5\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{-}}f(x)=1\)
\(\,\,\,\,\,\)The limit is \(1\)
\(\textbf{6)}\) Find \( \displaystyle \lim_{x\to 4^{+}} f(x) \)
where \(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
The limit is \( 1 \)
\(\,\,\,\,\,\text{Since }x\to4^{+}\text{, use the rule for }4 \lt x \lt 6.\)
\(\,\,\,\,\,f(x)=x-3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{+}}f(x)=4-3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{+}}f(x)=1\)
\(\,\,\,\,\,\)The limit is \(1\)
\(\textbf{7)}\) Find \( \displaystyle \lim_{x\to 6^{-}} f(x) \)
where \(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
The limit is \( 3 \)
\(\,\,\,\,\,\text{Since }x\to6^{-}\text{, use the rule for }4 \lt x \lt 6.\)
\(\,\,\,\,\,f(x)=x-3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to6^{-}}f(x)=6-3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to6^{-}}f(x)=3\)
\(\,\,\,\,\,\)The limit is \(3\)
\(\textbf{8)}\) Find \( \displaystyle \lim_{x\to 6^{+}} f(x) \)
where \(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
The limit is \( 6 \)
\(\,\,\,\,\,\text{Since }x\to6^{+}\text{, use the rule for }x\geq6.\)
\(\,\,\,\,\,f(x)=x\)
\(\,\,\,\,\,\displaystyle \lim_{x\to6^{+}}f(x)=6\)
\(\,\,\,\,\,\)The limit is \(6\)
\(\textbf{9)}\) Is the following continuous at \(x=4? \)
\(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
Yes, the piecewise function is continuous at \( x=4. \)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{-}}f(x)=-(4)+5=1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{+}}f(x)=4-3=1\)
\(\,\,\,\,\,f(4)=-(4)+5=1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to4^{-}}f(x)=\lim_{x\to4^{+}}f(x)=f(4)\)
\(\,\,\,\,\,\)Yes, the piecewise function is continuous at \(x=4\).
\(\textbf{10)}\) Is the following continuous at \( x=6? \)
\(f(x) = \begin{cases}
-x+5 & \text{if } x\leq 4 \\
x-3 & \text{if } 4\lt x \lt 6 \\
x & \text{if }x\geq 6
\end{cases}\)
No, the piecewise function is not continuous at \( x=6. \)
\(\,\,\,\,\,\displaystyle \lim_{x\to6^{-}}f(x)=6-3=3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to6^{+}}f(x)=6\)
\(\,\,\,\,\,3\neq6\)
\(\,\,\,\,\,\text{Since the one-sided limits are not equal, the function is not continuous at }x=6.\)
\(\,\,\,\,\,\)No, the piecewise function is not continuous at \(x=6\).
\(\textbf{11)}\) Find the value of a that makes the function continuous.
\(f(x) = \begin{cases}
\frac{x^2+6x+5}{x+1} & \text{if } x \neq -1 \\
a & \text{if }x = -1
\end{cases}\)
The answer is \( a=4 \)
\(\,\,\,\,\,\text{For continuity at }x=-1\text{, set }a=\displaystyle\lim_{x\to-1}\frac{x^2+6x+5}{x+1}.\)
\(\,\,\,\,\,\displaystyle \frac{x^2+6x+5}{x+1}=\frac{(x+1)(x+5)}{x+1}\)
\(\,\,\,\,\,\displaystyle \frac{x^2+6x+5}{x+1}=x+5\text{ when }x\neq-1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to-1}(x+5)=4\)
\(\,\,\,\,\,a=4\)
\(\,\,\,\,\,\)The answer is \(a=4\)
\(\textbf{12)}\) Find the values of a and b that make the function continuous at all points.
\(f(x) = \begin{cases}
2x^2 & \text{if } x\leq 2 \\
ax+b & \text{if } 2\lt x \lt 4 \\
x^2+4 & \text{if }x\geq 4
\end{cases}\)
The answer is \( a=6, b=-4 \)
\(\,\,\,\,\,\text{At }x=2\text{, the pieces must match.}\)
\(\,\,\,\,\,2(2)^2=2a+b\)
\(\,\,\,\,\,8=2a+b\)
\(\,\,\,\,\,\text{At }x=4\text{, the pieces must match.}\)
\(\,\,\,\,\,4a+b=4^2+4\)
\(\,\,\,\,\,4a+b=20\)
\(\,\,\,\,\,\text{Subtract }2a+b=8\text{ from }4a+b=20.\)
\(\,\,\,\,\,2a=12\)
\(\,\,\,\,\,a=6\)
\(\,\,\,\,\,2(6)+b=8\)
\(\,\,\,\,\,b=-4\)
\(\,\,\,\,\,\)The answer is \(a=6, b=-4\)
\(\textbf{13)}\) Find the value of k that makes the function continuous at all points.
\(f(x) = \begin{cases}
\sin{x} & \text{if } x\leq \pi \\
x-k & \text{if } x\geq \pi
\end{cases}\)
The answer is \( k=\pi \)
\(\,\,\,\,\,\displaystyle \lim_{x\to \pi^{-}} f(x) = \lim_{x\to \pi^{+}} f(x)\)
\(\,\,\,\,\, \sin{\pi} = \pi-k\)
\(\,\,\,\,\, 0 = \pi-k\)
\(\,\,\,\,\, k = \pi\)
\(\,\,\,\,\,\)The answer is \(k=\pi\)
\(\textbf{14)}\) Find \( \displaystyle \lim_{x\to 3^{-}} f(x) \)
where \(f(x)=\begin{cases}
2x-1 & \text{if }x \lt 3\\
x^2 & \text{if }x\geq3
\end{cases}\)
The limit is \(5\)
\(\,\,\,\,\,\text{Since }x\to3^{-}\text{, use the rule for }x \lt 3.\)
\(\,\,\,\,\,f(x)=2x-1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{-}}f(x)=2(3)-1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{-}}f(x)=5\)
\(\,\,\,\,\,\)The limit is \(5\)
\(\textbf{15)}\) Find \( \displaystyle \lim_{x\to 3^{+}} f(x) \)
where \(f(x)=\begin{cases}
2x-1 & \text{if }x \lt 3\\
x^2 & \text{if }x\geq3
\end{cases}\)
The limit is \(9\)
\(\,\,\,\,\,\text{Since }x\to3^{+}\text{, use the rule for }x\geq3.\)
\(\,\,\,\,\,f(x)=x^2\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{+}}f(x)=3^2\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{+}}f(x)=9\)
\(\,\,\,\,\,\)The limit is \(9\)
\(\textbf{16)}\) Is the following continuous at \(x=3?\)
\(f(x)=\begin{cases}
2x-1 & \text{if }x \lt 3\\
x^2 & \text{if }x\geq3
\end{cases}\)
No, the function is not continuous at \(x=3\).
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{-}}f(x)=2(3)-1=5\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3^{+}}f(x)=3^2=9\)
\(\,\,\,\,\,5\neq9\)
\(\,\,\,\,\,\text{Since the one-sided limits are not equal, the function is not continuous at }x=3.\)
\(\,\,\,\,\,\)No, the function is not continuous at \(x=3\).
\(\textbf{17)}\) Find \(a\) so that the function is continuous at \(x=5\).
\(f(x)=\begin{cases}
3x+a & \text{if }x \lt 5\\
2x^2-1 & \text{if }x\geq5
\end{cases}\)
The answer is \(a=34\)
\(\,\,\,\,\,\text{For continuity, the left and right expressions must match at }x=5.\)
\(\,\,\,\,\,3(5)+a=2(5)^2-1\)
\(\,\,\,\,\,15+a=50-1\)
\(\,\,\,\,\,15+a=49\)
\(\,\,\,\,\,a=34\)
\(\,\,\,\,\,\)The answer is \(a=34\)
\(\textbf{18)}\) Find \(k\) so that the function is continuous at \(x=1\).
\(f(x)=\begin{cases}
kx+2 & \text{if }x \lt 1\\
x^2+5 & \text{if }x\geq1
\end{cases}\)
The answer is \(k=4\)
\(\,\,\,\,\,\text{For continuity, the left and right expressions must match at }x=1.\)
\(\,\,\,\,\,k(1)+2=1^2+5\)
\(\,\,\,\,\,k+2=6\)
\(\,\,\,\,\,k=4\)
\(\,\,\,\,\,\)The answer is \(k=4\)
\(\textbf{19)}\) Find the values of \(a\) and \(b\) that make the function continuous at all points.
\(f(x)=\begin{cases}
x+2 & \text{if }x\leq1\\
ax+b & \text{if }1 \lt x \lt 3\\
2x^2-1 & \text{if }x\geq3
\end{cases}\)
The answer is \(a=7, b=-4\)
\(\,\,\,\,\,\text{At }x=1\text{, the pieces must match.}\)
\(\,\,\,\,\,1+2=a(1)+b\)
\(\,\,\,\,\,3=a+b\)
\(\,\,\,\,\,\text{At }x=3\text{, the pieces must match.}\)
\(\,\,\,\,\,3a+b=2(3)^2-1\)
\(\,\,\,\,\,3a+b=17\)
\(\,\,\,\,\,\text{Subtract }a+b=3\text{ from }3a+b=17.\)
\(\,\,\,\,\,2a=14\)
\(\,\,\,\,\,a=7\)
\(\,\,\,\,\,7+b=3\)
\(\,\,\,\,\,b=-4\)
\(\,\,\,\,\,\)The answer is \(a=7, b=-4\)
\(\textbf{20)}\) Find the value of \(c\) that makes the function continuous.
\(f(x)=\begin{cases}
\frac{x^2-9}{x-3} & \text{if }x\neq3\\
c & \text{if }x=3
\end{cases}\)
The answer is \(c=6\)
\(\,\,\,\,\,\text{For continuity at }x=3\text{, set }c=\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}.\)
\(\,\,\,\,\,\displaystyle \frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}\)
\(\,\,\,\,\,\displaystyle \frac{x^2-9}{x-3}=x+3\text{ when }x\neq3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to3}(x+3)=6\)
\(\,\,\,\,\,c=6\)
\(\,\,\,\,\,\)The answer is \(c=6\)
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