Chapter 2.2, Problem 65E

(a)

To determine

To find: The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{1}{x}$ , $g\left(x\right)=x$ and $c=0$ .

Answer to Problem 65E

The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{1}{x}$ , $g\left(x\right)=x$ and $c=0$ doesn’t exist, $0$ and $1$ respectively.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{1}{x}$ , $g\left(x\right)=x$ and $c=0$ .

Calculation:

Evaluate the limit.

  $\underset{x\to 0^{-}}{\lim }\left(\frac{1}{x}\right)=-\infty$

And,

  $\underset{x\to 0^{+}}{\lim }\left(\frac{1}{x}\right)=\infty$

So, the limit of function $f\left(x\right)=\frac{1}{x}$ doesn’t exist at $x=0$ as $\underset{x\to 0^{-}}{\lim }\left(\frac{1}{x}\right)\ne \underset{x\to 0^{+}}{\lim }\left(\frac{1}{x}\right)$ .

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }g\left(x\right)=\underset{x\to 0}{\lim }x\\ =0\end{array}$

The limit of function $g\left(x\right)=x$ is $0$ .

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

Evaluate the product function.

  $\begin{array}{c}fg=\frac{1}{x}\cdot x\\ =1\end{array}$

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }fg=\underset{x\to 0}{\lim }1\\ =1\end{array}$

So, the value of $\underset{x\to 0}{\lim }fg$ is equal to $1$ .

Therefore, the limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{1}{x}$ , $g\left(x\right)=x$ and $c=0$ doesn’t exist, $0$ and $1$ respectively.

(b)

To determine

To find: The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=-\frac{2}{x^3}$ , $g\left(x\right)=4x^3$ and $c=0$ .

Answer to Problem 65E

The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=-\frac{2}{x^3}$ , $g\left(x\right)=4x^3$ and $c=0$ doesn’t exist, $0$ and $-8$ respectively.

Explanation of Solution

Given information:

The functions $f\left(x\right)=-\frac{2}{x^3}$ , $g\left(x\right)=4x^3$ and $c=0$ .

Calculation:

Evaluate the limit.

  $\underset{x\to 0^{-}}{\lim }\left(-\frac{2}{x^3}\right)=-\infty$

And,

  $\underset{x\to 0^{+}}{\lim }\left(-\frac{2}{x^3}\right)=\infty$

So, the limit of function $f\left(x\right)=-\frac{2}{x^3}$ doesn’t exist at $x=0$ as $\underset{x\to 0^{-}}{\lim }\left(-\frac{2}{x^3}\right)\ne \underset{x\to 0^{+}}{\lim }\left(-\frac{2}{x^3}\right)$ .

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }g\left(x\right)=\underset{x\to 0}{\lim }4x^3\\ =4\underset{x\to 0}{\lim }{x}^3\\ =0\end{array}$

The limit of function $g\left(x\right)=4x^3$ is $0$ .

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

Evaluate the product function.

  $\begin{array}{c}fg=-\frac{2}{x^3}\cdot 4x^3\\ =-8\end{array}$

Evaluate the limit.

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

So, the value of $\underset{x\to 0}{\lim }fg$ is equal to $-8$ .

Therefore, the limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=-\frac{2}{x^3}$ , $g\left(x\right)=4x^3$ and $c=0$ doesn’t exist, $0$ and $-8$ respectively.

(c)

To determine

To find: The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{3}{x-2}$ , $g\left(x\right)={\left(x-2\right)}^3$ and $c=2$ .

Answer to Problem 65E

The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{3}{x-2}$ , $g\left(x\right)={\left(x-2\right)}^3$ and $c=2$ doesn’t exist, $0$ and $0$ respectively.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3}{x-2}$ , $g\left(x\right)={\left(x-2\right)}^3$ and $c=0$ .

Calculation:

Evaluate the limit.

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

And,

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

So, the limit of function $f\left(x\right)=\frac{3}{x-2}$ doesn’t exist at $x=0$ as $\underset{x\to 2^{-}}{\lim }\left(\frac{3}{x-2}\right)\ne \underset{x\to 2^{+}}{\lim }\left(\frac{3}{x-2}\right)$ .

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 2}{\lim }g\left(x\right)=\underset{x\to 2}{\lim }{\left(x-2\right)}^3\\ ={\left(\underset{x\to 2}{\lim }\left(x-2\right)\right)}^3\\ =0\end{array}$

So, the value of $\underset{x\to 2}{\lim }g\left(x\right)$ is equal to $0$ .

Evaluate the product function.

  $\begin{array}{c}fg=\frac{3}{x-2}\cdot {\left(x-2\right)}^3\\ =3{\left(x-2\right)}^2\end{array}$

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 2}{\lim }fg=\underset{x\to 2}{\lim }\left(3{\left(x-2\right)}^2\right)\\ =3\underset{x\to 2}{\lim }\left({\left(x-2\right)}^2\right)\\ =3\times 0\\ =0\end{array}$

So, the value of $\underset{x\to 2}{\lim }fg$ is equal to $0$ .

Therefore, the limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{3}{x-2}$ , $g\left(x\right)={\left(x-2\right)}^3$ and $c=2$ doesn’t exist, $0$ and $0$ respectively.

(d)

To determine

To find: The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{5}{{\left(3-x\right)}^4}$ , $g\left(x\right)={\left(x-3\right)}^2$ and $c=3$ .

Answer to Problem 65E

The limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{5}{{\left(3-x\right)}^4}$ , $g\left(x\right)={\left(x-3\right)}^2$ and $c=3$ doesn’t exist, $0$ and $1$ respectively.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{5}{{\left(3-x\right)}^4}$ , $g\left(x\right)={\left(x-3\right)}^2$ and $c=3$ .

Calculation:

Evaluate the limit.

  $\underset{x\to 3^{-}}{\lim }\left(\frac{5}{{\left(3-x\right)}^4}\right)=\infty$

And,

  $\underset{x\to 3^{+}}{\lim }\left(\frac{5}{{\left(3-x\right)}^4}\right)=\infty$

So, the value of $\underset{x\to 3}{\lim }f\left(x\right)$ is equal to $\infty$ .

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 3}{\lim }g\left(x\right)=\underset{x\to 3}{\lim }{\left(x-3\right)}^2\\ =0\end{array}$

So, the value of $\underset{x\to 3}{\lim }g\left(x\right)$ is equal to $0$ .

Evaluate the product function.

  $\begin{array}{c}fg=\frac{5}{{\left(3-x\right)}^4}\cdot {\left(x-3\right)}^2\\ =\frac{5}{{\left(x-3\right)}^4}\cdot {\left(x-3\right)}^2\\ =\frac{5}{{\left(x-3\right)}^2}\end{array}$

Evaluate the limit.

  $\begin{array}{c}\underset{x\to 3}{\lim }fg=\underset{x\to 3}{\lim }\frac{5}{{\left(x-3\right)}^2}\\ =\infty \end{array}$

So, the value of $\underset{x\to 3}{\lim }fg$ is equal to $\infty$ .

Therefore, the limits of $f$ , $g$ and $fg$ as for $f\left(x\right)=\frac{5}{{\left(3-x\right)}^4}$ , $g\left(x\right)={\left(x-3\right)}^2$ and $c=3$$\infty$ , $0$ and $\infty$ respectively.

(e)

To determine

To find: The value of $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)$ .

Answer to Problem 65E

The value of $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)$ can’t be determined.

Explanation of Solution

Given information:

Suppose that $\underset{x\to c}{\lim }f\left(x\right)=0$ and $\underset{x\to c}{\lim }g\left(x\right)=\infty$ .

Calculation:

Evaluate the limit.

  $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)=\underset{x\to c}{\lim }\left(f\left(x\right)\right)\cdot \underset{x\to c}{\lim }\left(g\left(x\right)\right)$

As we have seen in above subparts that sometimes the limit of product exist and sometimes not.

So, the value of $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)$ can’t be determined.