Question

determine the following limit, using ∞ or −∞ when appropriate, or state that it does not exist. (limlimits_{x \to 3^+} \frac{x^2 - 6x + 8}{x - 3}) select the correct choice below and, if necessary, fill in the answer box to complete your choice. (\bigcirc) a. (limlimits_{x \to 3^+} \frac{x^2 - 6x + 8}{x - 3} = square) (simplify your answer.) (\bigcirc) b. the limit does not exist and is neither (infty) nor (-infty).

Explanation:

Step1: Analyze the denominator as \(x \to 3^+\)

As \(x \to 3^+\), \(x - 3\) approaches \(0\) from the positive side, so \(x - 3\to 0^+\).

Step2: Analyze the numerator at \(x = 3\)

Substitute \(x = 3\) into the numerator: \(3^2 - 6\times3 + 8 = 9 - 18 + 8 = -1\). So the numerator approaches \(-1\) (a non - zero constant) as \(x \to 3^+\).

Step3: Determine the sign of the fraction

We have a fraction \(\frac{\text{negative constant}}{\text{positive small number}}\) as \(x \to 3^+\). So \(\frac{x^{2}-6x + 8}{x - 3}\to\frac{-1}{0^+}\), which means the limit is \(-\infty\).

Answer:

A. \(\lim\limits_{x\to 3^{+}}\frac{x^{2}-6x + 8}{x - 3}=-\infty\)