<s>[INST] Limit with missing variables 
Find the values for $a$ and $b$ such that 
  $$\lim_{x \to 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = 3$$

Basically what I did so far was I started by multiplying by the conjugate. and obtained $$\frac{a+bx-3}{x(\sqrt{a+bx}+\sqrt 3)}$$ I don't know what to do after this. 
 [/INST] $$ \lim_{x \to 0}\frac{\sqrt{a+bx}-\sqrt{3}}{x}=3 \Longrightarrow  \lim_{x \to 0}x\cdot\frac{\sqrt{a+bx}-\sqrt{3}}{x}=0 \cdot3=0$$
$$ i.e. \quad\lim_{x \to 0} (\sqrt{a+bx}-\sqrt{3})=0 \quad \therefore \sqrt{a}=\sqrt{3} ,\quad a=3$$
then 
$$ \lim_{x \to 0}\frac{\sqrt{3+bx}-\sqrt{3}}{x}=\lim_{x \to 0}\frac{(3+bx)-3}{x(\sqrt{3+bx}+\sqrt{3})}=\lim_{x \to 0}\frac{b}{(\sqrt{3+bx}+\sqrt{3})}=3 $$
We can find that
$$ \frac{b}{2\sqrt{3}}=3 , \quad b=2\sqrt{3}. $$
This is the answer.
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