Chapter 13, Problem 2T

For the piecewise-defined function f whose graph is shown, find:

1. (a) $\underset{x\to -1^{-}}{\lim }f(x)$
2. (b) $\underset{x\to -1^{+}}{\lim }f(x)$
3. (c) $\underset{x\to -1}{\lim }f(x)$
4. (d) $\underset{x\to -0^{-}}{\lim }f(x)$
5. (e) $\underset{x\to -0^{+}}{\lim }f(x)$
6. (f) $\underset{x\to 0}{\lim }f(x)$
7. (g) $\underset{x\to 2^{-}}{\lim }f(x)$
8. (h) $\underset{x\to 2^{+}}{\lim }f(x)$
9. (i) $\underset{x\to 2}{\lim }f(x)$

$f(x)=\{\begin{array}{ll}1\hfill & \text{if}x<-1\hfill \\ 0\hfill & \text{if}x=-1\hfill \\ x^2\hfill & \text{if}-1<x\le 2\hfill \\ 4-x\hfill & \text{if}2<x\hfill \end{array}$

To determine

To find: The value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ is 1.

Explanation of Solution

Given:

The graph of function f is given below,

Figure (1)

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 1 as x approaches $-1$.

For the limit $\underset{x\to -1^{-}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 1 when x goes to $-1$.

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

Thus, the value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ is 1.

To determine

To find: The value of $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ is 1.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 1 as x approaches $-1$.

For the limit $\underset{x\to -1^{+}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 1 when x goes to $-1$.

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

Thus the value of $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ is 1.

To determine

To find: The value of $\underset{x\to -1}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to -1}{\lim }f\left(x\right)$ is 1.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

If $\underset{x\to a^{-}}{\lim }f\left(x\right)=L$ and $\underset{x\to a^{+}}{\lim }f\left(x\right)=L$, then $\underset{x\to a}{\lim }f\left(x\right)$ is equal to the L.

From part (a), value of $\underset{x\to -1^{-}}{\lim }f\left(x\right)$ is 1 and from part (b) the value of $\underset{x\to -1^{+}}{\lim }f\left(x\right)$ is 1

$\underset{x\to -1}{\lim }f\left(x\right)=1$

Thus, the value of $\underset{x\to -1}{\lim }f\left(x\right)$ is equal to 1.

To determine

To find: The value of $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ is 0.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 0 as x approaches 0.

For the limit $\underset{x\to 0^{-}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 0 when x goes to 0.

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

Thus the value of $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ is 0.

To determine

To find: The value of $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ is 0.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 0 as x approaches 0.

For the limit $\underset{x\to 0^{+}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 0 when x goes to 0.

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

Thus the value of $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ is 0.

To determine

To find: The value of $\underset{x\to 0}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 0}{\lim }f\left(x\right)$ is 0.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

If $\underset{x\to a^{-}}{\lim }f\left(x\right)=L$ and $\underset{x\to a^{+}}{\lim }f\left(x\right)=L$, then $\underset{x\to a}{\lim }f\left(x\right)$ is equal to the L.

From part (d), the value of $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ is 0 and from part (e) the value of $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ is 0.

$\underset{x\to 0}{\lim }f\left(x\right)=0$

Thus, the value of $\underset{x\to 0}{\lim }f\left(x\right)$ is equal to 0.

To determine

To find: The value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ is 4.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 4 as x approaches 2.

For the limit $\underset{x\to 2^{-}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 4 when x goes to 2.

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=4$

Thus the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ is 4.

To determine

To find: The value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ is 2.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

From the Figure (1), the value of $f\left(x\right)$ approaches 2 as x approaches 2.

For the limit $\underset{x\to 2^{+}}{\lim }f\left(x\right)$, the value of $f\left(x\right)$ is 2 when x goes to 2.

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=2$

Thus, the value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ is 2.

To determine

To find: The value of $\underset{x\to 2}{\lim }f\left(x\right)$ and the graph of f is given and the value of function $f\left(x\right)$ is $f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Answer to Problem 2T

The value of $\underset{x\to 2}{\lim }f\left(x\right)$ does not exist.

Explanation of Solution

Given:

The graph of function f is given in Figure (1).

The value of function $f\left(x\right)$ is,

$f\left(x\right)=\{\begin{array}{l}1\text{ if }x<-1\\ 0\text{ if }x=-1\\ x^2\text{ if }-1<x\le 2\\ 4-x\text{ if 2}<x\end{array}$

Calculation:

If $\underset{x\to a^{-}}{\lim }f\left(x\right)=L$ and $\underset{x\to a^{+}}{\lim }f\left(x\right)=L$, then $\underset{x\to a}{\lim }f\left(x\right)$ is equal to the L.

From part (g), the value $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ of is 4 and from part (h) the value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ is 2.

$\underset{x\to 2^{-}}{\lim }f\left(x\right)\ne \underset{x\to 2^{+}}{\lim }f\left(x\right)$

Thus, the value of $\underset{x\to 2}{\lim }f\left(x\right)$ is does not exist.