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} ) b. does ( limlimits_{x \to 3} f(x) ) exist? if so, what is it? if not, why not? ( \bigcirc ) a. no, ( limlimits_{x \to 3} f(x) ) does not exist because ( f(3) ) is undefined. ( \bigcirc ) b. no, ( limlimits_{x \to 3} f(x) ) does not exist because ( limlimits_{x \to 3^+} f(x)
eq limlimits_{x \to 3^-} f(x) ). ( \bigcirc ) c. yes, ( limlimits_{x \to 3} f(x) ) exists and equals ( square ). (simplify your answer.) ( \bigcirc ) 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) ).

Explanation:

Step1: Find left - hand limit

For \(x\to3^{-}\), we use the function \(f(x)=5 - x\) (since \(x<3\)).
The left - hand limit \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}(5 - x)\).
Substitute \(x = 3\) into \(5 - x\): \(5-3 = 2\).

Step2: Find right - hand limit

For \(x\to3^{+}\), we use the function \(f(x)=\frac{2x}{3}\) (since \(x > 3\)).
The right - hand limit \(\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}\frac{2x}{3}\).
Substitute \(x = 3\) into \(\frac{2x}{3}\): \(\frac{2\times3}{3}=2\).

Step3: Determine if the limit exists

Since \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{+}}f(x) = 2\), by the definition of the limit of a function at a point, \(\lim_{x\to3}f(x)\) exists and is equal to \(2\). So the correct option is C, and the limit value is \(2\).

Answer:

C. Yes, \(\lim\limits_{x\to3}f(x)\) exists and equals \(2\).