Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can  the function $f(x)$  be considered as  $x^2 + 3$  and go about evaluating the limit using the Limit laws?


"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!
