Question

let
f(x)=\begin{cases}18&\text{if }x < - 7\\-x + 11&\text{if }-7leq x < 8\\1&\text{if }x = 8\\11&\text{if }x>8end{cases}
sketch the graph of this function and find the following limits, if they exist. (if a limit does not exist, enter dne.)

1. (lim_{x

ightarrow - 7^{-}}f(x)=)

1. (lim_{x

ightarrow - 7^{+}}f(x)=)

1. (lim_{x

ightarrow - 7}f(x)=)

1. (lim_{x

ightarrow8^{-}}f(x)=)

1. (lim_{x

ightarrow8^{+}}f(x)=)

1. (lim_{x

ightarrow8}f(x)=)

Explanation:

Step1: Left - hand limit at $x=-7$

Since $x\to - 7^{-}$, use $f(x)=18$. So $\lim_{x\to - 7^{-}}f(x)=18$.

Step2: Right - hand limit at $x=-7$

Since $x\to - 7^{+}$, use $f(x)=-x + 11$. $\lim_{x\to - 7^{+}}f(x)=-(-7)+11 = 18$.

Step3: Limit at $x=-7$

Since $\lim_{x\to - 7^{-}}f(x)=\lim_{x\to - 7^{+}}f(x)=18$, $\lim_{x\to - 7}f(x)=18$.

Step4: Left - hand limit at $x = 8$

Since $x\to 8^{-}$, use $f(x)=-x + 11$. $\lim_{x\to 8^{-}}f(x)=-8 + 11=3$.

Step5: Right - hand limit at $x = 8$

Since $x\to 8^{+}$, use $f(x)=11$. $\lim_{x\to 8^{+}}f(x)=11$.

Step6: Limit at $x = 8$

Since $\lim_{x\to 8^{-}}f(x)
eq\lim_{x\to 8^{+}}f(x)$, $\lim_{x\to 8}f(x)=\text{DNE}$.

Answer:

1. $18$
2. $18$
3. $18$
4. $3$
5. $11$
6. DNE