Question

use the following function and its graph to answer parts a through d below. let ( f(x) = \begin{cases} 8 - x, & x < 6 \\ 3, & x = 6 \\ \frac{x}{3}, & x > 6 end{cases} ) ... b. does ( limlimits_{x \to 6} f(x) ) exist? if so, what is it? if not, why not? (\bigcirc) a. yes, ( limlimits_{x \to 6} f(x) ) exists and equals (square). (simplify your answer.) (\bigcirc) b. no, ( limlimits_{x \to 6} f(x) ) does not exist because ( f(6) ) is undefined. (\bigcirc) c. no, ( limlimits_{x \to 6} f(x) ) does not exist because ( f(6) ) is not equal to ( limlimits_{x \to 6^+} f(x) ) or ( limlimits_{x \to 6^-} f(x) ). (\bigcirc) d. no, ( limlimits_{x \to 6} f(x) ) does not exist because ( limlimits_{x \to 6^+} f(x)
eq limlimits_{x \to 6^-} f(x) ).

Explanation:

Step1: Find left - hand limit

For \(x\to6^{-}\), we use the function \(f(x)=8 - x\) (since \(x<6\) for left - hand limit). Substitute \(x = 6\) into \(8 - x\): \(8-6 = 2\). So, \(\lim_{x\to6^{-}}f(x)=2\).

Step2: Find right - hand limit

For \(x\to6^{+}\), we use the function \(f(x)=\frac{x}{3}\) (since \(x > 6\) for right - hand limit). Substitute \(x = 6\) into \(\frac{x}{3}\): \(\frac{6}{3}=2\).

Step3: Determine existence of limit

Since \(\lim_{x\to6^{-}}f(x)=\lim_{x\to6^{+}}f(x) = 2\), the limit \(\lim_{x\to6}f(x)\) exists and is equal to 2.

Answer:

2