Question

(4) 11. find and classify the discontinuities for the function
g(x)=\begin{cases}\frac{x^{2}-3x}{x^{3}-2x^{2}-3x}&\text{if }x
eq0\\1&\text{if }x = 0end{cases}
a) list any removable discontinuities:
b) list any jump discontinuities:
c) list any infinite discontinuities:

Explanation:

Step1: Simplify the function for $x

eq0$
First, factor the numerator and denominator of $\frac{x^{2}-3x}{x^{3}-2x^{2}-3x}$. Numerator: $x^{2}-3x=x(x - 3)$. Denominator: $x^{3}-2x^{2}-3x=x(x^{2}-2x - 3)=x(x - 3)(x+1)$. So, for $x
eq0$, $g(x)=\frac{x(x - 3)}{x(x - 3)(x + 1)}=\frac{1}{x + 1}$.

Step2: Check the limit as $x

ightarrow0$
Find $\lim_{x
ightarrow0}g(x)$. Since for $x
eq0$, $g(x)=\frac{1}{x + 1}$, then $\lim_{x
ightarrow0}g(x)=\lim_{x
ightarrow0}\frac{1}{x + 1}=1$. And $g(0)=1$. So, there is no discontinuity at $x = 0$.

Step3: Check for other discontinuities

Set the denominator of the simplified - function $y=\frac{1}{x + 1}$ equal to zero. $x+1 = 0$ gives $x=-1$. $\lim_{x
ightarrow - 1^{-}}\frac{1}{x + 1}=-\infty$ and $\lim_{x
ightarrow - 1^{+}}\frac{1}{x + 1}=\infty$.

Answer:

a) None
b) None
c) $x=-1$