Q. 11 - Calculus

Question

Each function in Exercises 9–12 is discontinuous at some value x = c. Describe the type of discontinuity and any one-sided continuity at x = c, and sketch a possible graph of f. 

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1, \underset{x\to 0^{+}}{\lim }f\left(x\right)=1, f\left(0\right)=0.$

Step-by-Step Solution

Answer

The type of discontinuity is a jump discontinuity and there is not any one-sided continuity.

The possible graph of f is

Step 1. Given Information.

The given function is $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1, \underset{x\to 0^{+}}{\lim }f\left(x\right)=1, f\left(0\right)=0.$

Step 2. Describing the discontinuity.

From the function, we can depict that $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1 \mathrm{and} \underset{\mathrm{x}\to 0^{+}}{\lim }f\left(x\right)=1$ both exist but are not equal to $f\left(0\right)=0.$

Thus, f(x) has a jump discontinuity at $x=0.$

Step 3. Describing one-sided continuity at x=c .

There is not any one-sided continuity at $x=0$ because  $\underset{x\to 0^{-}}{\lim }\ne f\left(0\right) \mathrm{and} \underset{\mathrm{x}\to 0^{+}}{\lim }\ne \mathrm{f}\left(0\right).$

Step 4. Graph of f .

The graph of f is