Evaluate the given limit

"Can the function $f(x)$ be considered as $x^2 + 3$ and go about solving the limit using the Limit laws?"

No, since we have only that $|f(x)-3|\le x^2\implies 3-x^2\le f(x)\le 3+x^2$.

But we can proceed by using $\color{blue}{f(x)-3=O(x^2)}$, where we are using the ("Big O notation").

Then, we can evaluate the limit of interest by writing

$$\begin{align} \frac{f(x)-\sqrt{x^2+9}}{x}&=\frac{f(x)-3\left(1+\frac{x^2}{9}\right)^{1/2}}{x}\\\\ &=\frac{f(x)-3\left(1+\color{red}{\frac12 \frac{x^2}{9}+O(x^4)}\right)}{x}\\\\ &=\frac{\color{blue}{\left(f(x)-3\right)}+\color{red}{O(x^2)}}{x}\\\\ &=\frac{\color{blue}{O(x^2)}+\color{red}{O(x^2)}}{x}\\\\ &=O(x)\to 0\,\,\text{as}\,\,x\to 0 \end{align}$$

And we are done!