Question

1. (5 points) (0.5 point each) use the graph to calculate the following limits.

a) $lim_{x
ightarrow - 6^{-}}f(x)=$
b) $lim_{x
ightarrow - 6^{+}}f(x)=$
c) $lim_{x
ightarrow - 6}f(x)=$
d) $f(-6)=$
e) $lim_{x
ightarrow3}f(x)=$
f) $lim_{x
ightarrow3^{+}}f(x)=$
g) $lim_{x
ightarrow3}f(x)=$
h) $f(3)=$
i) $lim_{x
ightarrow6}f(x)=$
j) $f(6)=$

Explanation:

Step1: Analyze left - hand limit as $x\to - 6$

As $x$ approaches $-6$ from the left side ($x\to - 6^{-}$), we look at the values of the function on the graph to the left of $x = - 6$. The function approaches a $y$-value of $2$. So, $\lim_{x\to - 6^{-}}f(x)=2$.

Step2: Analyze right - hand limit as $x\to - 6$

As $x$ approaches $-6$ from the right side ($x\to - 6^{+}$), we look at the values of the function on the graph to the right of $x=-6$. The function approaches a $y$-value of $2$. So, $\lim_{x\to - 6^{+}}f(x)=2$.

Step3: Analyze limit as $x\to - 6$

Since $\lim_{x\to - 6^{-}}f(x)=\lim_{x\to - 6^{+}}f(x) = 2$, then $\lim_{x\to - 6}f(x)=2$.

Step4: Find the function value at $x=-6$

The open - circle at $x = - 6$ means the function is not defined at this point in the normal sense. But if we assume the function value is given by the filled - in dot above $x=-6$, $f(-6)$ is not equal to the limit. However, if we go by the graph's main trend, we note that the function is not continuous at $x=-6$ and we assume the value based on the filled - in dot, $f(-6)$ is not relevant to the limit calculation and from the graph, we can't precisely say its value from the limit perspective. But if we consider the filled - in dot, $f(-6)=5$.

Step5: Analyze left - hand limit as $x\to3$

As $x$ approaches $3$ from the left side ($x\to3^{-}$), the function approaches a $y$-value of $- \infty$ because of the vertical asymptote. So, $\lim_{x\to3^{-}}f(x)=-\infty$.

Step6: Analyze right - hand limit as $x\to3$

As $x$ approaches $3$ from the right side ($x\to3^{+}$), the function approaches a $y$-value of $+\infty$. So, $\lim_{x\to3^{+}}f(x)=+\infty$.

Step7: Analyze limit as $x\to3$

Since $\lim_{x\to3^{-}}f(x)
eq\lim_{x\to3^{+}}f(x)$, $\lim_{x\to3}f(x)$ does not exist.

Step8: Find the function value at $x = 3$

The function is not defined at $x = 3$ (vertical asymptote), so $f(3)$ is undefined.

Step9: Analyze limit as $x\to6$

As $x$ approaches $6$ from both the left and the right, the function approaches a $y$-value of $3$. So, $\lim_{x\to6}f(x)=3$.

Step10: Find the function value at $x = 6$

The open - circle at $x = 6$ means $f(6)$ is not defined.

Answer:

a) $\lim_{x\to - 6^{-}}f(x)=2$
b) $\lim_{x\to - 6^{+}}f(x)=2$
c) $\lim_{x\to - 6}f(x)=2$
d) $f(-6)=5$ (assuming the filled - in dot value)
e) $\lim_{x\to3^{-}}f(x)=-\infty$
f) $\lim_{x\to3^{+}}f(x)=+\infty$
g) $\lim_{x\to3}f(x)$ does not exist
h) $f(3)$ is undefined
i) $\lim_{x\to6}f(x)=3$
j) $f(6)$ is undefined