Chapter 1, Problem 1RE

Classify each statement as either true or false.

Match each function in column A with the most appropriate rule to use for differentiating the function. $\left[1.5,1.6\right]$

If $\underset{x\to -5}{\lim }f\left(x\right)$ exists, then $f\left(5\right)$ must exist. $\left[1.1\right]$

To determine

Whether the statement “If $\underset{x\to -5}{\lim }f\left(x\right)$ exists, then $f\left(5\right)$ must exist” is true or false.

Answer to Problem 1RE

Solution:

The statement “If $\underset{x\to -5}{\lim }f\left(x\right)$ exists, then $f\left(5\right)$ must exist” is false.

Explanation of Solution

Consider the limit $\underset{x\to -5}{\lim }f\left(x\right)$.

Assume $f\left(x\right)=\frac{1}{x-5}$. Therefore,

$\begin{array}{ccc}\hfill \underset{x\to -5}{\lim }f\left(x\right)& =& \underset{x\to -5}{\lim }\left(\frac{1}{x-5}\right)\hfill \\ \hfill & =& \frac{1}{\left(-5\right)-5}\hfill \\ \hfill & =& -\frac{1}{10}\hfill \end{array}$

Thus, the limit $\underset{x\to -5}{\lim }f\left(x\right)$ for the function $f\left(x\right)=\frac{1}{x-5}$ exists.

Now, the value of $f\left(5\right)$ is,

$\begin{array}{ccc}\hfill f\left(5\right)& =& \frac{1}{5-5}\hfill \\ \hfill & =& \frac{1}{0}\hfill \end{array}$

As $f\left(5\right)=\frac{1}{0}$ which is not defined. So, $f\left(5\right)$ does not exist.

Therefore, $\underset{x\to -5}{\lim }f\left(x\right)$ exists but $f\left(5\right)$ not exist.

Hence, the provided statement is false.