Question

use the following function and its graph to answer parts a through d below. let ( f(x) = \begin{cases} 5 - x, & x < 3 \\ 3, & x = 3 \\ dfrac{2x}{3}, & x > 3 end{cases} ) graph shown. c. yes, ( limlimits_{x \to 3} f(x) ) exists and equals 2. (simplify your answer.) d. no, ( limlimits_{x \to 3} f(x) ) does not exist because ( f(3) ) is not equal to ( limlimits_{x \to 3^+} f(x) ) or ( limlimits_{x \to 3^-} f(x) ). c. find ( limlimits_{x \to 2^+} f(x) ). select the correct choice below and, if necessary, fill in the answer box in your choice. a. ( limlimits_{x \to 2^+} f(x) = square ) (simplify your answer.) b. the limit does not exist.

Explanation:

Step1: Determine the relevant function

For \( x \to 2^+ \), we check the domain of the piecewise function. Since \( 2^+ \) means \( x \) is slightly greater than 2, and \( 2 < 3 \), we use the function \( f(x)=5 - x \) (because the first piece \( 5 - x \) is defined for \( x < 3 \), and \( 2^+ \) is still less than 3).

Step2: Evaluate the limit

To find \( \lim_{x \to 2^+} f(x) \), we substitute \( x = 2 \) into the function \( f(x)=5 - x \) (since the function is continuous on \( x < 3 \), the right - hand limit as \( x \to 2 \) is just the value of the function at \( x = 2 \) for this piece).
So, \( \lim_{x \to 2^+}(5 - x)=5 - 2 \).
Calculating \( 5 - 2 = 3 \).

Answer:

A. \( \lim_{x \to 2^+} f(x)=\boxed{3} \)