What is $\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$ when it exists? 
For what values of $a$ and $b$ does the following limit exist and what is the limit in those cases?
  $$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$

Actually this is an assignment which I need to do, but I have no idea where to start, please help!

A more brute-force approach:
$$|x+3| = \begin{cases}
x + 3, & x \geq -3 \\
-(x+3) = -x - 3, & x < -3 
\end{cases}$$
so
$$\dfrac{|x+3|(\sqrt {ax+b}-2)}{x} = \begin{cases}
\dfrac{(x+3)(\sqrt {ax+b}-2)}{x},  & x \geq -3 \\
\dfrac{(-x-3)(\sqrt {ax+b}-2)}{x},  & x < -3 \text{.}
\end{cases}$$
For $x$ within a "small" neighborhood of $0$, we need only worry about the $x \geq -3$ case:
$$\begin{align}
\dfrac{(x+3)(\sqrt {ax+b}-2)}{x} &= (x+3)\dfrac{\sqrt {ax+b}-2}{x}\text{.}
\end{align}$$
We're not splitting the fraction since obviously $$\lim_{x \to 0}\dfrac{2}{x}$$
does not exist. 
As $x \to 0$, $x + 3 \to 3$. 
Being clever, we could notice that if the limit of $\dfrac{\sqrt {ax+b}-2}{x}$ exists as $x \to 0$, this can be written as
$$\lim_{x \to 0}\dfrac{\sqrt {ax+b}-2}{x} = \lim_{x \to 0}\dfrac{f(x)-f(0)}{x-0} = f^{\prime}(0)$$
where $f(x) = \sqrt{ax+b}$, and $f(0) = \sqrt{b} = 2$, so $b = 4$. Furthermore, $$f^{\prime}(x) = \dfrac{a}{2\sqrt{ax+b}} = \dfrac{a}{2\sqrt{ax+4}}\text{.}$$
We then have 
$$f^{\prime}(0) = \dfrac{a}{2\sqrt{4}} = \dfrac{a}{4}\text{.}$$
Thus,
$$\lim_{x \to 0}\dfrac{|x+3|(\sqrt {ax+b}-2)}{x} = \dfrac{3a}{4}$$
as long as $b = 4$. Any $a$ will do.
Furthermore, this relies on the existence of $$\lim_{x \to 0}\dfrac{\sqrt {ax+b}-2}{x}$$
perhaps there is a way to relax this assumption. 
