Chapter 12.2, Problem 33E

a.

To determine

To find: To approximate the limit by using a graphing utility to graph the function.

Answer to Problem 33E

The approximate limit by using a graphing utility to graph the given function is $-0.125$ .

Explanation of Solution

Given:$\underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}$

  

The above graph is the graphical representation of the given function with the limits $\underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}$ . From the graph we observe the limit $-0.125$ .

Thus we can find the limit using graphing utility.

b.

To determine

To find: To numerically approximate the limit by using the table feature of the graphing utility to create a table.

Answer to Problem 33E

The numerically approximate limit using a table from the graph is $-0.125$ .

Explanation of Solution

Given:$\underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}$

From the graph we can create a table,

x	f(x)
16.5	-0.124038
16.4	-0.124228
16.3	-0.124419
16.2	-0.12612
16.1	-0.124805
16	Indeterminate
15.9	-0.125196

Therefore we can conclude that numerically the limit of the given function does is $-0.125$ .

c.

To determine

To find: To algebraically evaluate the limit by using the appropriate techniques.

Answer to Problem 33E

By evaluating the given limit algebraically the limit is $-0.125$ .

Explanation of Solution

Given:$\underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}$

Initially factor the numerator and denominator,

  $\begin{array}{l}\underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}=\underset{x\to 16}{\lim }\frac{4-\sqrt{x}}{x-16}\cdot \frac{4+\sqrt{x}}{4+\sqrt{x}}\\ \underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}=\underset{x\to 16}{\lim }\frac{16-x}{\left(x-16\right)\left(4+\sqrt{x}\right)}\\ \underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}=\underset{x\to 16}{\lim }\frac{-1}{4+\sqrt{x}},\text{ by using direct substitution}\\ \underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}=-\frac{1}{8}\\ \underset{x\to {16}^{+}}{\lim }\frac{4-\sqrt{x}}{x-16}=-0.125\end{array}$

Thus we can approximate the limit algebraically.