Chapter 2.4, Problem 37E

b.

To determine

To check:whether the given function f has a removable discontinuity at a . if the discontinuity is removable, then find a function g that agrees with f for $x\ne a$ and is continuous ata .

Answer to Problem 37E

Removable discontinuity.

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{x^4-1}{x-1}a=1$ .

Calculation:

The given function is $\begin{array}{l}f\left(x\right)=\frac{x^4-1}{x-1}=\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{x-1}=\left(x+1\right)\left(x^2+1\right)=x^3+x^2+x+1\\ =\end{array}$ .

f can be extended to a continuous function on $ℝ$ includingat $x=1$ .

Define $g\left(x\right)=x^3+x^2+x+1$

Then g is defined and continuities for all real numbers and $g\left(x\right)=f\left(x\right),x\ne 2$

Therefore, $g\left(x\right)=x^3+x^2+x+1$ is removable discontinuity.

b.

To determine

To check: whether the given function f has a removable discontinuity at a . if the discontinuity is removable, then find a function g that agrees with f for $x\ne a$ and is continuous at a .

Answer to Problem 37E

Removable discontinuity.

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{x^2-x-2}{x-2}a=2$ .

Calculation:

The given function is $f\left(x\right)=\frac{x^2-x-2}{x-2}=x\left(x+1\right)$ .

f can be extended to a continuous function on $ℝ$ including $x=2$ .

Define $g\left(x\right)=x\left(x+1\right)$

Then g is defined and continuities for all real numbers and $g\left(x\right)=f\left(x\right),x\ne 2$

Therefore, $g\left(x\right)=x\left(x+1\right)$ is removable discontinuity

c.

To determine

To check: whether the given function f has a removable discontinuity at a . if the discontinuity is removable, then find a function g that agrees with f for $x\ne a$ and is continuous at a .

Answer to Problem 37E

Jump discontinuity and non-removable.

Explanation of Solution

Given:

The given function is $f\left(x\right)=\left[\sin x\right]a=\pi$ .

Calculation:

The given function is $f\left(x\right)=\left[\sin x\right]a=\pi$ .

Remember that x is the “greatest integer” function, which equals the largest not integer not greater than x .

  $\begin{array}{l}\left[\left[2\right]\right]=2\\ \left[\left[2.3\right]\right]=2\\ \left[\left[-0.2\right]\right]=-1\end{array}$

For $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.<x<\pi$ , then

  $0<\sin x<1\text{ and }\left[\left[\sin x\right]\right]=0$ , so

  $\underset{x\to {\pi }^{-}}{\lim }\left[\left[\sin x\right]\right]=0$

TFor $\pi <x<\raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ , then

  $-1<\sin x<0\text{ and }\left[\left[\sin x\right]\right]=-1$ , so

  $\underset{x\to {\pi }^{+}}{\lim }\left[\left[\sin x\right]\right]=-1$

Thus, $\underset{x\to {\pi }^{-}}{\lim }\left[\left[\sin x\right]\right]\ne \underset{x\to {\pi }^{+}}{\lim }\left[\left[\sin x\right]\right]$

  $\left[\left[\sin x\right]\right]$ is a step function, so it is a jump discontinuity and non-removable.