Chapter 1.2, Problem 65E

(a)

To determine

To find: The limit of $f,g$ and $f_g$ as $x\to c$ .

Answer to Problem 65E

  $\underset{x\to 0}{\lim }f(x)$ = does not exists.

  $\underset{x\to 0}{\lim }g(x)=0$

  $\underset{x\to 0}{\lim }[f(x)g(x)]=1$

Explanation of Solution

Given information: The functions are:

  $f(x)=\frac{1}{x}$

  $g(x)=x$

  $c=0$

Calculation:

Note that, $\underset{x\to 0^{-}}{\lim }\frac{1}{x}=-\infty$

And

  $\underset{x\to 0^{+}}{\lim }\frac{1}{x}=\infty$

Therefore, $\underset{x\to 0}{\lim }f(x)$ = does not exists.

(b)

To determine

To find: The limit of $f,g$ and $f_g$ as $x\to c$ .

Answer to Problem 65E

  $\begin{array}{c}f\to \infty \text{ as }x\to 0^{-}\\ f\to -\infty \text{ as }x\to 0^{+}\\ g\to 0\text{ as }x\to 0\\ fg\to -8\text{ as }x\to 0\end{array}$

Explanation of Solution

Given information: The functions are:

  $f(x)=-\frac{2}{x^2}$

  $g(x)=4x^3$

  $c=0$

Calculation:

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }\left(-\frac{2}{x^3}\right)\\ =-\frac{2}{{(\text{ positive })}^3}\\ =-\infty \end{array}$

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }\left(-\frac{2}{x^3}\right)\\ =-\frac{2}{{\text{ (negative) }}^3}\\ =\infty \end{array}$

Therefore, $f\to \infty \text{ as }x\to 0^{-}$ and

  $f\to -\infty \text{ as }x\to 0^{+}$

  $\underset{x\to 0}{\lim }g(x)=\underset{x\to 0}{\lim }4x^3$

  $=4{(0)}^3$

  $=0$

Therefore, $g\to 0\text{ as }x\to 0$

  $fg=-\frac{2}{x^3}\cdot 4x^3=-8$

  $\begin{array}{c}\underset{x\to }{\lim }fg=\underset{x\to 0}{\lim }(-8)\\ =-8\end{array}$

Hence,, $fg\to -8\text{ as }x\to 0$ .

Therefore the required limit of $f,g$ and $f_g$ as $x\to c$ are:

  $\begin{array}{c}f\to \infty \text{ as }x\to 0^{-}\\ f\to -\infty \text{ as }x\to 0^{+}\\ g\to 0\text{ as }x\to 0\\ fg\to -8\text{ as }x\to 0\end{array}$

(c)

To determine

To find: The limit of $f,g$ and $f_g$ as $x\to c$ .

Answer to Problem 65E

  $\underset{x\to 2}{\lim }f(x)\text{ does not exist}\text{. }$

  $\underset{x\to 2}{\lim }g(x)=0$

  $\underset{x\to 2}{\lim }[f(x)g(x)]=0$

Explanation of Solution

Given information: The functions are:

  $f(x)=\frac{3}{x-2}$

  $g(x)={(x-2)}^3$

  $c=2$

Calculation:

Note that,

  $\underset{x\to 2^{-}}{\lim }\frac{3}{x-2}=-\infty$

And,

  $\underset{x\to 2^{+}}{\lim }\frac{3}{x-2}=\infty$

Therefore, $\underset{x\to 2}{\lim }f(x)\text{ does not exist}\text{. }$

As g , put $x=0$ :

  $\underset{x\to 0}{\lim }g(x)={(2-2)}^3$

  $=0$

The limit of $fg$ as:

  $\underset{x\to 2}{\lim }[f(x)g(x)]=\underset{x\to 2}{\lim }\left[\frac{3}{x-2}\cdot {(x-2)}^3\right]$

  $=\underset{x\to 2}{\lim }3{(x-2)}^2$

  $=0$

Therefore, the required limit of $f,g$ and $f_g$ as $x\to c$ are:

  $\underset{x\to 2}{\lim }f(x)\text{ does not exist}\text{. }$

  $\underset{x\to 2}{\lim }g(x)=0$

  $\underset{x\to 2}{\lim }[f(x)g(x)]=0$

(d)

To determine

To find: The limit of $f,g$ and $f_g$ as $x\to c$ .

  $\begin{array}{c}\underset{x\to 3}{\lim }f(x)=\infty \\ \underset{x\to 3}{\lim }g(x)=0\\ \underset{x\to 3}{\lim }f(x)g(x)=\infty \end{array}$

Answer to Problem 65E

Explanation of Solution

Given information: The functions are:

  $f(x)=\frac{5}{{(3-x)}^4}$

  $g(x)={(x-3)}^2$

  $c=3$

Calculation:

  $\begin{array}{c}\underset{x\to 3}{\lim }f(x)=\underset{x\to 3}{\lim }\frac{5}{{(3-x)}^4}\\ =\infty \end{array}$

And,

  $\underset{x\to 3}{\lim }g(x)=\underset{x\to 3}{\lim }{(x-3)}^2$

  $={(3-3)}^2$

  $=0$

And,

  $\underset{x\to 3}{\lim }f(x)g(x)=\underset{x\to 3}{\lim }\frac{5}{{(3-x)}^4}\cdot {(x-3)}^2$

  $=\underset{x\to 3}{\lim }\frac{5}{{(3-x)}^2}$

  $=\infty$

Therefore the required limit of $f,g$ and $f_g$ as $x\to c$ are:

limit of $f,g$ and $f_g$ as $x\to c$ .

  $\begin{array}{c}\underset{x\to 3}{\lim }f(x)=\infty \\ \underset{x\to 3}{\lim }g(x)=0\\ \underset{x\to 3}{\lim }f(x)g(x)=\infty \end{array}$

(e)

To determine

To find: The explanation of $\underset{x\to c}{\lim }(f(x)\cdot g(x))$

No conclusion can be reached.

Answer to Problem 65E

Explanation of Solution

Given information: The functions are:

  $\underset{x\to c}{\lim }f(x)=0$ and

  $\underset{x\to c}{\lim }g(x)=\infty$

Calculation:

Part (a), (b) and (c) $x=c$ is a vertical asymptote where the lateral limits are different directions $(\infty \text{ and }-\infty )$ .

The limits of $f$ (a)-(c) is infinites limits.

The limit $fg$ may be finite or infinite.

Therefore, the required no conclusion can be reached.