(a) Estimate the value of $\displaystyle \lim \limits_{x \to 0} \frac{\sqrt{3+x}-\sqrt{3}}{x}$ by using the graph of $f(x)$
Based on the graph, the limit of $f(x)$ seems to have a value of 0.29 as $x$ approaches 0.
(b) Guess the limit the limit by using a table of values of $f(x)$
$\begin{array}{|c|c|c|c|c|}
\hline\\
x & 0.01 & 0.02 & 0.03 & 0.04 \\ \hline
f(x) & 0.2884 & 0.2882 & 0.2879 & 0.2877\\
\hline
\end{array} $
Based on the values from the table, the limit of the function seems to have a value of 0.29 as $x$ approaches to 0.
(c) Find the exact value of the limit using the limit laws.
$
\begin{equation}
\begin{aligned}
& \lim\limits_{x \to 0} \frac{\sqrt{3 + x} - \sqrt{3}}{x}
\cdot
\frac{\sqrt{3 + x} + \sqrt{3}}
{\sqrt{3 + x} + \sqrt{3}}
= \lim \limits_{x \to 0} \frac{3 + x - 3}{ x( \sqrt{3 + x} + \sqrt{3} ) }
&& \text{ Simplify the equation by multiplying both numerator and denominator by } \sqrt{3 + x} + \sqrt{3}
\\
&\lim \limits_{x \to 0} \frac{1}{\sqrt{3 + x} + \sqrt{3}}
= \frac{\lim \limits_{x \to 0} 1 }{\lim \limits_{x \to 0} \sqrt{3 + x} + \lim \limits_{x \to 0} \sqrt{3}}
&& \text{ Quotient and root law.}
\\
& \lim \limits_{x \to 0}\frac{1}{\sqrt{3 + x} + \sqrt{3}}
= \frac{1}{\sqrt{\lim \limits_{x \to 0}(3 + x)} + \sqrt{3}}
&& \text{ Constant and root law.}\\
& \lim \limits_{x \to 0} \frac{1}{\sqrt{3 + x} + \sqrt{3}}
= \frac{1 }{\sqrt{3 + 0} + \sqrt{3}}
&& \text{ Sum, constant and special limit law.}\\
\end{aligned}
\end{equation}\\
\boxed{\displaystyle \lim \limits_{x \to 0} \frac{1}{\sqrt{3 + x} + \sqrt{3}} = \frac{1}{2 \sqrt{3}}}
$
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