Chapter 12, Problem 33RE

To determine

Tofind:The value of the limit $\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}$ and verify the result graphically.

Answer to Problem 33RE

The value of the limit $\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}$ is $-\frac{1}{8}$ .

Explanation of Solution

Given information:

The given limit is $\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}$ .

Calculation:

In the given equation, the value of $u$ gives a zero value in the denominator. So, eliminate the denominator.

  $\begin{array}{c}\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}=\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}\times \left(\frac{\sqrt{16-u}+4}{\sqrt{16-u}+4}\right)\\ =\underset{u\to 0}{\lim }\frac{16-u-16}{u\left(\sqrt{16-u}+4\right)}\\ =\underset{u\to 0}{\lim }\frac{-u}{u\left(\sqrt{16-u}+4\right)}\\ =\underset{u\to 0}{\lim }\frac{-1}{\sqrt{16-u}+4}\end{array}$

Substitute the limit and solve.

  $\begin{array}{c}\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}=-\frac{1}{\sqrt{16-u}+4}\\ =-\frac{1}{\sqrt{16}+4}\\ =-\frac{1}{4+4}\\ =-\frac{1}{8}\end{array}$

Hence, the value of the limit is $-\frac{1}{8}$ .

Verify the result graphically:

Draw the graph for $\underset{u\to 0}{\lim }\frac{\sqrt{16-u}-4}{u}$ using graphical utility.

  

From the graph, it is clear that the value approaches to $-\frac{1}{8}$ for $u=0$ . Hence, proved.