Chapter 11.3, Problem 88E

Solution

a.

The limits of $f$ , $g$ and $fg$ as $x$ approaches $c$ .

The limits of the functions $f$ , $g$ and $fg$ as $x$ approaches $c=0$ are $\infty$ , $0$ and $2$ respectively.

Given information:

  $f(x)=\frac{2}{x^2},g(x)=x^2,c=0$

Calculation:

Find the function $fg$ .

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

Now, find the limits.

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

Conclusion:

The limits of the functions $f$ , $g$ and $fg$ as $x$ approaches $c=0$ are $\infty$ , $0$ and $2$ respectively.

b.

The limits of $f$ , $g$ and $fg$ as $x$ approaches $c$ .

The limits are $\underset{x\to 0}{\lim }f(x)=\infty$ , $\underset{x\to 0}{\lim }g(x)=0$ and $\underset{x\to 0}{\lim }\left(fg\right)\left(x\right)$ does not exist.

Given information:

  $f(x)=\left|\frac{1}{x}\right|,g(x)=\sqrt[3]{x},c=0$

Calculation:

Find the function $fg$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left|\frac{1}{x}\right|\cdot \sqrt[3]{x}\\ =\frac{\sqrt[3]{x}}{\left|x\right|}\end{array}$

Now, find the limits.

  $\begin{array}{c}\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\left(\left|\frac{1}{x}\right|\right)\\ =\infty \\ \underset{x\to 0}{\lim }g(x)=\underset{x\to 0}{\lim }\left(\sqrt[3]{x}\right)\\ =0\end{array}$

The limit $\underset{x\to 0}{\lim }\left(fg\right)\left(x\right)$ does not exist as $\underset{x\to 0^{-}}{\lim }\left(fg\right)\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{\sqrt[3]{x}}{\left|x\right|}\right)=-\infty$ and $\underset{x\to 0^{+}}{\lim }\left(fg\right)\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{\sqrt[3]{x}}{\left|x\right|}\right)=\infty$ .

Conclusion:

The limits are $\underset{x\to 0}{\lim }f(x)=\infty$ , $\underset{x\to 0}{\lim }g(x)=0$ and $\underset{x\to 0}{\lim }\left(fg\right)\left(x\right)$ does not exist.

c.

The limits of $f$ , $g$ and $fg$ as $x$ approaches $c$ .

The limits are $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=0$ .

Given information:

  $f(x)=\left|\frac{3}{x-1}\right|,g(x)={(x-1)}^2,c=1$

Calculation:

Find the function $fg$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ $ =\left|\frac{3}{x-1}\right|\cdot {(x-1)}^2$=\frac{3{(x-1)}^2}{\left|x-1\right|}\end{array}$

Now, find the limits.

  $\begin{array}{c}\underset{x\to 1}{\lim }f(x)=\underset{x\to 1}{\lim }\left(\left|\frac{3}{x-1}\right|\right)\\ \underset{x\to 1}{\lim }f(x)=\infty \\ \underset{x\to 1}{\lim }g(x)=\underset{x\to 1}{\lim }\left({(x-1)}^2\right)\\ \underset{x\to 1}{\lim }g(x)=0\end{array}$

The limit $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=0$ as $\underset{x\to 1^{-}}{\lim }\left(fg\right)\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(\frac{3{(x-1)}^2}{\left|x-1\right|}\right)=0$ and $\underset{x\to 1^{+}}{\lim }\left(fg\right)\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(\frac{3{(x-1)}^2}{\left|x-1\right|}\right)=0$ .

Conclusion:

The limits are $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=0$ .

d.

The limits of $f$ , $g$ and $fg$ as $x$ approaches $c$ .

The limits are $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=\infty$ .

Given information:

  $f(x)=\frac{1}{{(x-1)}^4},g(x)={(x-1)}^2,c=1$

Calculation:

Find the function $fg$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ $ =\frac{1}{{(x-1)}^4}\cdot {(x-1)}^2$=\frac{1}{{(x-1)}^2}\end{array}$

Now, find the limits.

  $\begin{array}{c}\underset{x\to 1}{\lim }f(x)=\underset{x\to 1}{\lim }\left(\frac{1}{{(x-1)}^4}\right)\\ \underset{x\to 1}{\lim }f(x)=\infty \\ \underset{x\to 1}{\lim }g(x)=\underset{x\to 1}{\lim }\left({(x-1)}^2\right)\\ \underset{x\to 1}{\lim }g(x)=0\\ \underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=\underset{x\to 1}{\lim }\left(\frac{1}{{(x-1)}^2}\right)=\infty \end{array}$

Conclusion:

The limits are $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=\infty$ .

e.

The tell what can be said about the limit $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)$ if $\underset{x\to c}{\lim }f\left(x\right)=\infty$ and $\underset{x\to c}{\lim }g\left(x\right)=0$ based on the result from part (a)-(d).

The limit of the product of the functions at a point does not depend on the individual limits of the function at that point.

Given information:

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

Explanation:

Observe that $\underset{x\to c}{\lim }f\left(x\right)=\infty$ , $\underset{x\to c}{\lim }g\left(x\right)=0$ and $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)=2$ in part (a).

In part (b), $\underset{x\to 0}{\lim }f(x)=\infty$ , $\underset{x\to 0}{\lim }g(x)=0$ and $\underset{x\to 0}{\lim }\left(fg\right)\left(x\right)$ does not exist.

In part (c), $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=0$ .

In part (d), $\underset{x\to 1}{\lim }f(x)=\infty$ , $\underset{x\to 1}{\lim }g(x)=0$ and $\underset{x\to 1}{\lim }\left(fg\right)\left(x\right)=\infty$ .

These results show that the limit of the product of the functions at a point does not depend on the individual limits of the function at that point.