Question

$\left(a\right).$Use a graph of $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$

to estimate the value of$\underset{x\to 0}{\lim }f\left(x\right)$ to two decimal places.

$\left(b\right).$Use a table of values of$f\left(x\right)$to estimate the limit to four decimal places.

$\left(c\right).$Use the Limit Laws to find the exact value of the limit.

Short answer

The value of limit of the given function is$\frac{1}{2\sqrt{3}}.$

Step-by-step solution

Step 1. Given information.

The function here given is,

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

$\left(a\right)$

Step 2. Rationalize numerator.

Given,$f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$

Then,$\begin{array}{l}\frac{\sqrt{3+x}-\sqrt{3}}{x}:\frac{1}{\sqrt{x+3}+\sqrt{3}}\\ =\underset{x\to 0}{\lim }\left(\frac{1}{\sqrt{x+3}+\sqrt{3}}\right)\end{array}$

Put in the value $x=0$

$\begin{array}{c}=\frac{1}{\sqrt{0+3}+\sqrt{3}}\\ =\frac{1}{2\sqrt{3}}\end{array}$

Step 3. Draw the graph.

Now, as a graphical verification use a graphing device.

From the graph it is clear that,$\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}=\frac{1}{2\sqrt{3}}$

$\left(b\right)$

Step 4. Make the table.

Now, make a table by putting the value of $x.$

On the basis of the values in the table we make the guess that,

$\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}=0.2886=\frac{1}{2\sqrt{3}}$

$\left(c\right)$

Step 5. Using the Limit Laws.

The function given,$\underset{x\to 0}{\lim }\left(\frac{1}{\sqrt{x+3}+\sqrt{3}}\right)$

Now use limit of a root we get,

$\begin{array}{c}\frac{\underset{x\to 0}{\lim }1}{\underset{x\to 0}{\lim }\sqrt{x+3}+\underset{x\to 0}{\lim }\sqrt{3}}\\ =\frac{1}{\sqrt{3}+\sqrt{3}}\\ =\frac{1}{2\sqrt{3}}\end{array}$

Therefore, the guess is correct that value of limit of the given function is$\frac{1}{2\sqrt{3}}.$