classify the discontinuities in the function $h(x)=\frac{x^{3}-7x^{2}+10x}{x(x - 8)}$ at the points $x = 0$…

Explanation:

Step1: Simplify the function

First, factor the numerator $x^{3}-7x^{2}+10x=x(x^{2}-7x + 10)=x(x - 2)(x - 5)$. So the function $h(x)=\frac{x(x - 2)(x - 5)}{x(x - 8)}$, and for $x\neq0$, we can cancel out the $x$ terms, getting $h(x)=\frac{(x - 2)(x - 5)}{x - 8}$.

Step2: Analyze the discontinuity at $x = 0$

We find the limit as $x\rightarrow0$. $\lim_{x\rightarrow0}h(x)=\lim_{x\rightarrow0}\frac{(x - 2)(x - 5)}{x - 8}=\frac{(0 - 2)(0 - 5)}{0 - 8}=-\frac{10}{8}=-\frac{5}{4}$. Since the limit exists as $x\rightarrow0$ but the function is not defined at $x = 0$, it has a removable discontinuity at $x=0$.

Step3: Analyze the discontinuity at $x = 8$

We find the limit as $x\rightarrow8$. $\lim_{x\rightarrow8}\frac{(x - 2)(x - 5)}{x - 8}$. As $x\rightarrow8$, the numerator approaches $(8 - 2)(8 - 5)=18$ and the denominator approaches $0$. $\lim_{x\rightarrow8^{+}}\frac{(x - 2)(x - 5)}{x - 8}=+\infty$ and $\lim_{x\rightarrow8^{-}}\frac{(x - 2)(x - 5)}{x - 8}=-\infty$. So it has an infinite discontinuity at $x = 8$.

Answer:

Removable discontinuity at $x = 0$; Infinite discontinuity at $x = 8$