Chapter 9.2, Problem 58E

a.

To determine

To show:

  $\underset{x\to 0}{\lim }\frac{f^{\text{'}}\left(x\right)}{g^{\text{'}}\left(x\right)}=1\text{ but }\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=2$

Explanation of Solution

Given information :

  $f\left(x\right)=\{\begin{array}{l}x+2x\ne 0\\ 0x=0\end{array}\text{ and }g\left(x\right)=\{\begin{array}{l}x+1x\ne 0\\ 0x=0\end{array}$

Formula used : L’Hospital Rule.

Calculation :

We have $f^{\text{'}}\left(x\right)=1\text{ and }g\left(x\right)=1$

Now,

  $\underset{x\to 0}{\lim }\frac{f^{\text{'}}\left(x\right)}{g^{\text{'}}\left(x\right)}=\frac{1}{1}\text{ =1 }$

And , $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{\lim }\frac{x+2}{x+1}=\frac{0+2}{0+1}=2$

b.

To determine

To explain: Why the above limits does not contradict L’Hospital Rule.

Explanation of Solution

Given information :

  $f\left(x\right)=\{\begin{array}{l}x+2x\ne 0\\ 0x=0\end{array}\text{ and }g\left(x\right)=\{\begin{array}{l}x+1x\ne 0\\ 0x=0\end{array}$

Concept used : L’Hospital Rule.

This does not contradict the L’Hospital rule.

  $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is not in indeterminate from since $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=2$ as $x\to 0$ (but not equal to $0$ ).