<s>[INST] How to set up equation to find the unknown values if a limit exist. 
For what values of the constants $a$ and $b$ does the following limit exist?
  $$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$

for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$
$$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$
Firstly, I found difficulties in setting up two equation solving for a and b.
Secondly, since $|x+3|=x+3$ only when $x\to0$, I can only use the follwoing to set up the equation. $$\lim_{x\to0}\frac{(x+3)(\sqrt{ax+b}-2)}x$$ 
However, I cannot simplified it to cancel out the denominator $x$ by some method like rationalization. It gets $$\lim_{x\to0}\frac{(x+3)(ax+b-4)}{(\sqrt{ax+b}-2)x}$$.
Therefore, how can we set up the equation to find $a$ and $b$ or it needs other methods to do this type of questions?
 [/INST] Note that we can write
$$\begin{align}
\frac{|x+3|(\sqrt{ax+b}-2)}{x}&=\left(\frac{|x+3|(\sqrt{ax+b}-2)}{x}\right)\left(\frac{(\sqrt{ax+b}+2)}{(\sqrt{ax+b}+2)}\right)\\\\
&=\frac{|x+3|(ax+b-4)}{x(\sqrt{ax+b}+2)}\\\\
&=\left(\frac{|x+3|}{(\sqrt{ax+b}+2)}\right)\left(a+\frac{b-4}{x}\right)\tag 1
\end{align}$$
The first parenthetical term on the right-hand side of $(1)$ approaches $\frac{3}{2+\sqrt b}$ as $x\to 0$.  The limit of the second term does not exist unless $b=4$.  If $b=4$, then that limit is $a$, and we find 
$$\lim_{x\to 0}\frac{|x+3|(\sqrt{ax+4}-2)}{x}=\frac{3a}{4}$$
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