Chapter 11.2, Problem 26E

To determine

To find: the value of the limit.

Answer to Problem 26E

The value of the limit is $\frac{1}{8}$

Explanation of Solution

Given:

  $\underset{x\to 2}{\lim }\frac{4-\sqrt{18-x}}{x-2}$

  $\begin{array}{l}\underset{x\to 2}{\lim }\frac{4-\sqrt{18-x}}{x-2}=\underset{x\to 2}{\lim }\frac{4-\sqrt{18-x}}{x-2}*\frac{4+\sqrt{18-x}}{4+\sqrt{18-x}}\\ =\underset{x\to 2}{\lim }\frac{16-\left(18-x\right)}{\left(x-2\right)\left(4+\sqrt{18-x}\right)}\\ =\underset{x\to 2}{\lim }\frac{x-2}{\left(x-2\right)\left(4+\sqrt{18-x}\right)}\\ =\underset{x\to 2}{\lim }\frac{1}{4+\sqrt{18-x}}\\ =\frac{1}{8}\left(By\text{ direct substitution}\right)\end{array}$

See this from the graph of the function:

  

Conclusion:

The value of the limit is $\frac{1}{8}$