Answer
$$\lim_{x \to 0} \frac{\sqrt{2x+9}-3}{x}=\frac{1}{3}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
x & -0.001 & -0.01 & -0.1 & 0 & 0.001 & 0.01 & 0.1 \\ \hline
f(x) & 0.33520605 & 0.33351872 & 0.33335185 & \text{undefined} & 0.33331482 & 0.33314835 & 0.33150178 \\ \hline
\end{array}$$
Work Step by Step
Looking at the graph, we can find that when $x$ approaches $0$ from left and right, the function approaches $\frac{1}{3}$. So we can conclude that $\lim_{x \to 0} \frac{\sqrt{2x+9}-3}{x}$ is approximately $\frac{1}{3}$. The table also confirmed our conclusion.
Now, we want to find the limit analytically. Since we get the indeterminate form $\frac{0}{0}$ by direct substitution, we should rationalize the numerator:$$\frac{\sqrt{2x+9}-3}{x}=\left (\frac{\sqrt{2x+9}-3}{x} \right ) \left ( \frac{\sqrt{2x+9}+3}{\sqrt{2x+9}+3} \right )=\frac{2x}{x(\sqrt{2x+9}+3)}=\frac{2}{\sqrt{2x+9}+3}$$(Please note that $x \neq 0$). Hence,$$\lim_{ x \to 0} \frac{\sqrt{2x+9}-3}{x}=\lim_{x \to 0} \frac{2}{\sqrt{2x+9}+3}=\frac{1}{3}.$$
