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} ) ... d. does ( limlimits_{x \to 2} f(x) ) exist? if so, what is it? if not, why not? select the correct choice below and, if necessary, fill in the answer box in your choice.
a. no, ( limlimits_{x \to 2} f(x) ) does not exist because ( limlimits_{x \to 2^+} f(x)
eq limlimits_{x \to 2^-} f(x) ).
b. no, ( limlimits_{x \to 2} f(x) ) does not exist because ( f(2) ) is not equal to ( limlimits_{x \to 2^+} f(x) ) or ( limlimits_{x \to 2^-} f(x) ).
c. no, ( limlimits_{x \to 2} f(x) ) does not exist because ( f(2) ) is undefined.
d. yes, ( limlimits_{x \to 2} f(x) ) exists and equals (square). (simplify your answer.)

Explanation:

Step1: Determine the function for \( x \to 2 \)

Since \( 2 < 6 \), we use the part of the piecewise function \( f(x)=8 - x \) for \( x < 6 \).

Step2: Evaluate the limit

To find \( \lim_{x \to 2} f(x) \), we substitute \( x = 2 \) into \( 8 - x \) (because the function is continuous here for \( x < 6 \), so the limit is the function value at \( x = 2 \) for this piece).
So, \( \lim_{x \to 2} (8 - x)=8 - 2 = 6 \).

Answer:

6