Chapter 2, Problem 5RQ

To determine

To find: Whether the statement “ If ${\lim }_{x\to 5}f\left(x\right)=0$ and ${\lim }_{x\to 5}g\left(x\right)=0$ , then ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]$ does not exist” is true or false.

Answer to Problem 5RQ

The statement “ If ${\lim }_{x\to 5}f\left(x\right)=0$ and ${\lim }_{x\to 5}g\left(x\right)=0$ , then ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]$ does not exist” is false.

Explanation of Solution

Given information:

The given statement is “If ${\lim }_{x\to 5}f\left(x\right)=0$ and ${\lim }_{x\to 5}g\left(x\right)=0$ , then ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]$ does not exist”.

Calculation:

Let consider ${\lim }_{x\to 5}f\left(x\right)=0$ and ${\lim }_{x\to 5}g\left(x\right)=0$ .

Consider the two functions as

  $f\left(x\right)=\sin \left(x-5\right)$

  $g\left(x\right)=x-5$

The expression is

  ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]$

Substitute the values in the above expression.

  $\begin{array}{c}{\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]=\underset{x\to 5}{\lim }\frac{\sin \left(x-5\right)}{x-5}\\ =1\end{array}$

  ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]=1$ does exist hence the statement is false.

Therefore, the statement “ If ${\lim }_{x\to 5}f\left(x\right)=0$ and ${\lim }_{x\to 5}g\left(x\right)=0$ , then ${\lim }_{x\to 5}\left[f\left(x\right)/g\left(x\right)\right]$ does not exist” is false.