Question

(1 point)
let
f(x)=\begin{cases}11&\text{if }x < - 1\\-x + 10&\text{if }-1leq x<8\\3&\text{if }x = 8\\10e^{x - 8}&\text{if }x>8end{cases}
sketch the graph of this function and find the following limits, if they exist.

1. (lim_{x

ightarrow - 1^{-}}f(x)=)

1. (lim_{x

ightarrow - 1^{+}}f(x)=)

1. (lim_{x

ightarrow - 1}f(x)=)

1. (lim_{x

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

1. (lim_{x

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

1. (lim_{x

ightarrow8}f(x)=)
note: type inf for (infty) and -inf for (-infty). if the limit does not exist in another way, write dne.
note: you can earn partial credit on this problem.
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Explanation:

Step1: Find $\lim_{x\to - 1^{-}}f(x)$

When $x\to - 1^{-}$, we use the part of the function $f(x)=11$ (since $x < - 1$ for this left - hand limit). So $\lim_{x\to - 1^{-}}f(x)=11$.

Step2: Find $\lim_{x\to - 1^{+}}f(x)$

When $x\to - 1^{+}$, we use the part of the function $f(x)=-x + 10$ (since $-1\leq x<8$). Substitute $x=-1$ into $-x + 10$, we get $-(-1)+10=11$. So $\lim_{x\to - 1^{+}}f(x)=11$.

Step3: Find $\lim_{x\to - 1}f(x)$

Since $\lim_{x\to - 1^{-}}f(x)=\lim_{x\to - 1^{+}}f(x)=11$, then $\lim_{x\to - 1}f(x)=11$.

Step4: Find $\lim_{x\to 8^{-}}f(x)$

When $x\to 8^{-}$, we use the part of the function $f(x)=-x + 10$ (since $-1\leq x<8$). Substitute $x = 8$ into $-x + 10$, we get $-8 + 10=2$. So $\lim_{x\to 8^{-}}f(x)=2$.

Step5: Find $\lim_{x\to 8^{+}}f(x)$

When $x\to 8^{+}$, we use the part of the function $f(x)=10e^{x - 8}$. Substitute $x = 8$ into $10e^{x - 8}$, we get $10e^{8 - 8}=10$. So $\lim_{x\to 8^{+}}f(x)=10$.

Step6: Find $\lim_{x\to 8}f(x)$

Since $\lim_{x\to 8^{-}}f(x)=2$ and $\lim_{x\to 8^{+}}f(x)=10$, and $\lim_{x\to 8^{-}}f(x)
eq\lim_{x\to 8^{+}}f(x)$, then $\lim_{x\to 8}f(x)$ does not exist (DNE).

Answer:

1. $11$
2. $11$
3. $11$
4. $2$
5. $10$
6. DNE