8 mark for review the function h is defined by h(x) = (x² - 7)/(x - 3). which of the following statements…

Explanation:

Step1: Analyze left - hand limit

Consider the left - hand limit as $x\to3^{-}$. Let $x = 3 - \epsilon$, where $\epsilon>0$ and $\epsilon\to0$. Then $h(x)=\frac{(3 - \epsilon)^{2}-7}{(3 - \epsilon)-3}=\frac{9-6\epsilon+\epsilon^{2}-7}{-\epsilon}=\frac{2-6\epsilon+\epsilon^{2}}{-\epsilon}$. As $\epsilon\to0$, the dominant term is $\frac{2}{-\epsilon}$, so $\lim_{x\to3^{-}}h(x)=-\infty$.

Step2: Analyze right - hand limit

Consider the right - hand limit as $x\to3^{+}$. Let $x = 3+\epsilon$, where $\epsilon>0$ and $\epsilon\to0$. Then $h(x)=\frac{(3 + \epsilon)^{2}-7}{(3+\epsilon)-3}=\frac{9 + 6\epsilon+\epsilon^{2}-7}{\epsilon}=\frac{2+6\epsilon+\epsilon^{2}}{\epsilon}$. As $\epsilon\to0$, the dominant term is $\frac{2}{\epsilon}$, so $\lim_{x\to3^{+}}h(x)=+\infty$.

Answer:

C. $\lim_{x\to3^{-}}h(x)=-\infty$ and $\lim_{x\to3^{+}}h(x)=+\infty$