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?
 A: 
"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!
A: We have $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}| \le x^2 + |3 - \sqrt{x^2 + 9}|$. So:
$$\left|\frac{f(x)-\sqrt{x^2+9}}{x}\right| \le |x| + \left| \frac{\sqrt{x^2+9}-3}{x}\right|$$
Now squeeze.
A: Apply law |a+b| <= |a| + |b|
Let $L = \frac{f(x)-\sqrt{x^2+9}}{x}$
$$
|L| = \lvert\frac{f(x) - 3 + 3 - \sqrt{x^2+9}}{x}\rvert \le |\frac{f(x)-3}{x}| + |\frac{3 - \sqrt{x^2+9}}{x}| \\
\le |\frac{x^2}{x}| + |\frac{(3-\sqrt{x^2+9})(3+\sqrt{x^2+9})}{x(3+\sqrt{x^2+9})}| = |x| + |\frac{x}{3+\sqrt{x^2+9}}|
$$
Hence,
$$
\lim_{x \rightarrow 0}{|L|} \le \lim_{x \rightarrow 0}{(|x| + |\frac{x}{3+\sqrt{x^2+9}}|)} = 0 \quad (1)
$$
Since $|L| \ge 0 \quad \forall x$,  we also have $\lim_{x \rightarrow 0}{|L|} \ge 0 \quad (2)$
From (1) and (2) we have $\lim_{x \rightarrow 0}{|L|} = 0$, or  $\quad \lim_{x \rightarrow 0}L = 0$
