Chapter 13.1, Problem 14E

To determine

To estimate: The value of the limit.

Answer to Problem 14E

The value of the limit is $0\cdot 167$

Explanation of Solution

Given information:

Limit is $\underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}$

Concept used:

Limit is in $\frac{0}{0}$ form it is called the determinate form then we can use L’ Hospital’s Rule.

Formula used:

  $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\text{ when }\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}\text{ (or) }\frac{\pm \infty }{\pm \infty }$

Calculation:

Limit is $\underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}$

  $\frac{\sqrt{0+9}-3}{x}=\frac{0}{0}\text{ form}$

Therefore,

  $f(x)=\sqrt{x+9}-3g\left(x\right)=x$

  $\therefore$ Differentiate of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{1}{2\sqrt{x+9}}g\text{'}\left(x\right)=1\\ \underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}=\underset{x\to 0}{\lim }\frac{\frac{1}{2\sqrt{x+9}}-3}{x}=\frac{1}{2\sqrt{0+9}}=\frac{1}{6}\\ =0\cdot 167\end{array}$

Verification:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}$ which is required to graph.

Step 4: Press GRAPH button to graph the function and adjust the windows according to the graph.

The graph is obtained as:

  

From graph, it can be observed that points near by $x=0$ y - coordinate is $0\cdot 167$ nearly.