review the graph of function f(x), which is defined for -6 ≤ x ≤ 8. which one - sided limit does not exist…

Explanation:

Step1: Analyze $\lim_{x\rightarrow - 6^{-}}f(x)$

Function is not defined for $x < - 6$, but left - hand limit is not required to consider values outside domain on the left - side. Here, as $x$ approaches $-6$ from the left (within the domain $-6\leq x\leq8$), we can consider the behavior of the function near $x = - 6$. Since the function is defined on the closed - interval starting at $x=-6$, we can find the left - hand limit.

Step2: Analyze $\lim_{x\rightarrow2^{+}}f(x)$

As $x$ approaches $2$ from the right, we can follow the graph of the function for $x>2$. The function is well - behaved in this region and we can determine the value that the function approaches as $x$ gets closer to $2$ from the right.

Step3: Analyze $\lim_{x\rightarrow2^{-}}f(x)$

As $x$ approaches $2$ from the left, we can follow the graph of the function for $x < 2$. The function is well - behaved in this region and we can determine the value that the function approaches as $x$ gets closer to $2$ from the left.

Step4: Analyze $\lim_{x\rightarrow - 6^{+}}f(x)$

The function is defined starting at $x=-6$. There are no values of $x$ in the domain of the function that are less than $-6$. By the definition of a right - hand limit $\lim_{x\rightarrow a^{+}}f(x)$ where $a=-6$, we need to consider values of $x$ such that $a<x$ and $x$ is close to $a$. Since there are no such values of $x$ in the domain of $f(x)$ that satisfy this condition for $a = - 6$, the right - hand limit $\lim_{x\rightarrow - 6^{+}}f(x)$ does not exist.

Answer:

$\lim_{x\rightarrow - 6^{+}}f(x)$