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?

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.
