Chapter 9.2, Problem 69E

a.

To determine

To find: $f\left(x\right)\text{ and }g\left(x\right)$

Answer to Problem 69E

  $\begin{array}{l}f\left(x\right)=3x\\ g\left(x\right)=x\end{array}$

Explanation of Solution

Given information :

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }g\left(x\right)=\infty \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3\end{array}$

Concept used :

L’Hospital Rule.

  $x\to a\Rightarrow \left(x-a\right)\ne 0$

Calculation :

  $\begin{array}{l}\text{let as consider }f\left(x\right)=3x\text{ and }g\left(x\right)=x\text{ then both }f\text{ and }g\text{ tens to }\infty \text{ as }x\to \infty \\ \text{Now,}\\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{3x}{x}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to \infty }{\lim }\frac{3}{1}=3\end{array}$

b.

To determine

To find: $f\left(x\right)\text{ and }g\left(x\right)$

Answer to Problem 69E

  $\begin{array}{l}f\left(x\right)=x\\ g\left(x\right)=x^2\end{array}$

Explanation of Solution

Given information :

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

Concept used :

L’Hospital Rule.

  $x\to a\Rightarrow \left(x-a\right)\ne 0$

Calculation:

  $\begin{array}{l}\text{let as consider }f\left(x\right)=x\text{ and }g\left(x\right)=x^2\text{ then both }f\text{ and }g\text{ tens to }\infty \text{ as }x\to \infty \\ \text{Now,}\\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x}{x^2}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to \infty }{\lim }\frac{1}{2x}=0\end{array}$.

c.

To determine

To find: $f\left(x\right)\text{ and }g\left(x\right)$

Answer to Problem 69E

  $\begin{array}{l}f\left(x\right)=x^2\\ g\left(x\right)=x\end{array}$

Explanation of Solution

Given information :

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }g\left(x\right)=\infty \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty \end{array}$

Concept used :

L’Hospital Rule.

  $x\to a\Rightarrow \left(x-a\right)\ne 0$

Calculation:

  $\begin{array}{l}\text{let as consider }f\left(x\right)=x^2\text{ and }g\left(x\right)=x\text{ then both }f\text{ and }g\text{ tens to }\infty \text{ as }x\to \infty \\ \text{Now,}\\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{x}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to \infty }{\lim }\frac{2x}{1}=\infty \end{array}$