Chapter 1.3, Problem 19E

a.

To determine

To find: The point of discontinuity.

Answer to Problem 19E

The point of discontinuity occurs at $x=2$ .

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{ll}3-x,\hfill & x<2\hfill \\ \frac{x}{2}+1,\hfill & x>2\hfill \end{array}\right.$

Calculation:

Find the point of discontinuity:

Left hand limit:

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\left(3-x\right)\\ =1\end{array}$

Right hand limit:

  $\begin{array}{c}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\left(\frac{x}{2}+1\right)\\ =2\end{array}$

Here, left- and right-hand limits are not equal. The two-sided limit does not exist at

  $x=2$ . So, the point of discontinuity occurs at $x=2$ .

b.

To determine

To find: Whether the function is removal or non-removal discontinuity.

Answer to Problem 19E

There is a discontinuity at $x=2$ and it is nonremovable.

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{ll}3-x,\hfill & x<2\hfill \\ \frac{x}{2}+1,\hfill & x>2\hfill \end{array}\right.$

Calculation:

Graph $f\left(x\right)=\left\{\begin{array}{ll}3-x,\hfill & x<2\hfill \\ \frac{x}{2}+1,\hfill & x>2\hfill \end{array}\right.$

Since both of the lines, include $x=2$ , use open circles for both lines.

  

From the graph above and from part $\left(a\right)$ , there is a discontinuity at $x=2$ .

It is a nonremovable discontinuity, since the discontinuity is a jump discontinuity.