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} ) 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. yes, ( limlimits_{x \to 2} f(x) ) exists and equals (square). (simplify your answer.) d. no, ( limlimits_{x \to 2} f(x) ) does not exist because ( f(2) ) is undefined.

Explanation:

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

Since \( 2 < 3 \), the relevant part of the piecewise function for \( x \) approaching 2 (from both left and right, as \( 2 < 3 \)) is \( f(x)=5 - x \).

Step2: Calculate the left - hand limit as \( x\to 2^{-} \)

The left - hand limit \( \lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{-}}(5 - x) \). Substitute \( x = 2 \) into \( 5 - x \): \( 5-2 = 3 \).

Step3: Calculate the right - hand limit as \( x\to 2^{+} \)

The right - hand limit \( \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}(5 - x) \) (because \( 2^{+}<3 \), so we still use \( f(x) = 5 - x \)). Substitute \( x = 2 \) into \( 5 - x \): \( 5 - 2=3 \).

Step4: Determine the existence of the limit

For a limit \( \lim_{x\to a}f(x) \) to exist, \( \lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x) \). Here, \( \lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x) = 3 \), so \( \lim_{x\to 2}f(x) \) exists and is equal to 3. Also, the value of \( f(2) \) (even though we are finding the limit, and the limit does not depend on the value of the function at the point, but here we can note that \( f(2)=5 - 2 = 3 \) as well, but the key for the limit existence is the equality of left and right limits)

Answer:

C. Yes, \( \lim\limits_{x\to 2} f(x) \) exists and equals \( \boldsymbol{3} \).