<s>[INST] 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?
 [/INST] 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  $\quad \lim_{x \rightarrow 0}L = 0$
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