<s>[INST] Limit of $\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$? How to find this limit? 
For $|f(x)-3|\le x^2$,
$$\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$$
Can I make $f(x)=x^2+3$, and then
$$\lim_{x\to0}\frac{x^2+3-\sqrt{x+9}}{x}$$
Using l'Hopital's, 
$$\lim_{x\to0}\frac{2x-\frac{1}{2\sqrt{x+9}}}{1} = -\frac{1}{6}$$
I'm not confident about this answer. How do I do this problem?
 [/INST] Note that $ -x^2 \leq (f(x) - 3) \leq x^2$, implying that $ 3-x^2 \leq f(x) \leq 3+x^2$ for all $x$. Hence, fixing $x$, we have that:
$$
\frac{3-x^2-\sqrt{x+9}}{x} \leq \frac{f(x) - \sqrt{x+9}}{x} \leq \frac{3+x^2 - \sqrt{x+9}}{x}
$$
Now, take the limit as $x \to 0$ on both sides of this inequality. Both the left and right sides converge to $\frac {-1}6$ as $x \to 0$ (you can do l'hopital for this).
Hence, the answer is $\frac{-1}{6}$ by the squeeze theorem, but not because you assumed $f$ to be a certain function, only using what was given in a clever manner. 
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