Question

1. determine all x - values where the function g(x) given below has a discontinuity and classify each as removable, jump, or infinite. justify your answers.

g(x)=\begin{cases}3x + 1& \text{if }xleq0\\frac{2}{x}& \text{if }0 < x < 1\sin(pi x)+1& \text{if }1leq x<\frac{3}{2}\arctan(x - \frac{3}{2})& \text{if }xgeq\frac{3}{2}end{cases}

Explanation:

Step1: Check continuity at $x = 0$

Left - hand limit: $\lim_{x
ightarrow0^{-}}g(x)=\lim_{x
ightarrow0^{-}}(3x + 1)=3(0)+1 = 1$. Right - hand limit: $\lim_{x
ightarrow0^{+}}g(x)=\lim_{x
ightarrow0^{+}}\frac{2}{x}=\infty$. Since the right - hand limit is infinite, there is an infinite discontinuity at $x = 0$.

Step2: Check continuity at $x = 1$

Left - hand limit: $\lim_{x
ightarrow1^{-}}g(x)=\lim_{x
ightarrow1^{-}}\frac{2}{x}=2$. Right - hand limit: $\lim_{x
ightarrow1^{+}}g(x)=\lim_{x
ightarrow1^{+}}(\sin(\pi x)+1)=\sin(\pi)+1 = 1$. Since $\lim_{x
ightarrow1^{-}}g(x)
eq\lim_{x
ightarrow1^{+}}g(x)$, there is a jump discontinuity at $x = 1$.

Step3: Check continuity at $x=\frac{3}{2}$

Left - hand limit: $\lim_{x
ightarrow\frac{3}{2}^{-}}g(x)=\lim_{x
ightarrow\frac{3}{2}^{-}}(\sin(\pi x)+1)=\sin(\frac{3\pi}{2})+1=-1 + 1=0$. Right - hand limit: $\lim_{x
ightarrow\frac{3}{2}^{+}}g(x)=\lim_{x
ightarrow\frac{3}{2}^{+}}\arctan(x-\frac{3}{2})=\arctan(0)=0$. Also, $g(\frac{3}{2})=\arctan(0)=0$. So, the function is continuous at $x=\frac{3}{2}$.

Answer:

There is an infinite discontinuity at $x = 0$ and a jump discontinuity at $x = 1$.