Chapter 2.2, Problem 54E

(a)

To determine

To find: The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to -\infty$ .

Answer to Problem 54E

The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to -\infty$ is $1$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

Calculation:

Consider the given function $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

For $x\to -\infty$ , the function is $f\left(x\right)=\frac{x-2}{x-1}$ . So,

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

Therefore, the value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to -\infty$ is $1$ .

(b)

To determine

To find:The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to \infty$ .

Answer to Problem 54E

The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to \infty$ is $0$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

Calculation:

Consider the given function $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

For $x\to \infty$ , the function is $f\left(x\right)=\frac{1}{x^2}$ . So,

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{1}{x^2}\right)\\ =0\end{array}$

Therefore, the value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to \infty$ is $0$ .

(c)

To determine

To find:The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{-}$ .

Answer to Problem 54E

The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{-}$ is $2$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

Calculation:

As $x$ approaches to $0$ from the left side, the function is $f\left(x\right)=\frac{x-2}{x-1}$ .

The value of limit is:

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{x-2}{x-1}\right)\\ =\frac{0-2}{0-1}\\ =2\end{array}$

Therefore, the value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{-}$ is $2$ .

(d)

To determine

To find:The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{+}$ .

Answer to Problem 54E

The value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{+}$ is $\infty$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ .

Calculation:

As $x$ approaches to $0$ from the right side, the function is $f\left(x\right)=\frac{1}{x^2}$ .

The value of limit is:

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{1}{x^2}\right)\\ =\frac{1}{0}\\ =\infty \end{array}$

Therefore, the value of limit of $f\left(x\right)=\{\begin{array}{l}\frac{x-2}{x-1},x\le 0\\ \frac{1}{x^2},x>0\end{array}$ as $x\to 0^{+}$ is $\infty$ .