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. 

Being finite, numerator $\to 0\,\Rightarrow\,a = 3,$ so it is $\,\displaystyle\lim_{x\to0}\dfrac{f(x)-f(0)}x = f'(0),\ \ \ f(x) = \sqrt{3+bx}$
