Chapter 1.3, Problem 3QR

a.

To determine

To find: The limit of the given function

Answer to Problem 3QR

The limit of the function $\underset{x\to 2^{-}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ is $1$ .

Explanation of Solution

Given: $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$

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

Calculation:

Find $\underset{x\to 2^{-}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ :

In the limit as $x$ approaches from the left side, so choose the function which is

defined at $x<2$

So, the function is $f\left(x\right)=x^2-4x+5$

Substitute $x=2$ in the above function:

  $\begin{array}{c}f\left(2\right)=2^2-4\left(2\right)+5\\ =4-8+5\\ =-4+5\\ =1\end{array}$

Thus, the value of the function $\underset{x\to 2^{-}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ is $1$ .

b.

To determine

To find: The limit of the given function

Answer to Problem 3QR

The limit of the function $\underset{x\to 2^{+}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ is $2$ .

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$

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

Calculation:

Find $\underset{x\to 2^{+}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ :

Here $x$ approaches from the right side so the function which is defined at $x>2$ .

So, $f\left(x\right)=4-x$

Substitute $x=2$ in the above function:

  $\begin{array}{c}f\left(2\right)=4-2\\ =2\end{array}$

Thus, the value of the function $\underset{x\to 2^{-}}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ is $2$ .

c.

To determine

To find: The limit of the given function

Answer to Problem 3QR

The limit of the function $\underset{x\to 2}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ does not exist.

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$

Find $\underset{x\to 2}{\lim }f\left(x\right)$

Calculation:

Find $\underset{x\to 2}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$

Here, left- and right-hand limits are not equal, So, the limit does not exist.

So, the value of function $\underset{x\to 2}{\lim }\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$ does not exist.

d.

To determine

To find: The limit of the given function

Answer to Problem 3QR

The limit of the function is $f\left(2\right)$ is $2$ .

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-4x+5,\hfill & x<2\hfill \\ 4-x,\hfill & x\ge 2\hfill \end{array}\right.$

Find $f\left(2\right)$:

Calculation:

To find the value of function at $x=2$ , so choose the function which is defined at

  $x=2$ . So, the function is $f\left(x\right)=4-x$

Substitute $x=2$ in the above function,

  $\begin{array}{c}f\left(2\right)=4-2\\ =2\end{array}$

Therefore, the value of the function $f\left(2\right)$ is $2$ .