Answer
When $a = 4$ and $b= 4$, $\lim\limits_{x \to 0} \frac{\sqrt {ax-b} -2}{x} = 1$
Work Step by Step
We see a term containing a square root which can cause trouble. To get rid of this term we must make a substitution.
Thus we let $w = \sqrt {ax+b}$
As $x \to 0$, $w \to \sqrt b$
This gives $\lim\limits_{x \to 0} \frac{\sqrt{ax+b} -2}{x} = \lim\limits_{w \to \sqrt b} \frac{w - 2}{x}$
Now there is an x which must be removed. So we solve for x using our previous equation:
$w = \sqrt {ax+b}$
$w^{2} = ax + b$
$ax + b = w^{2}$
$ax = w^{2} - b$
$x = \frac{w^{2} - b}{a}$
Plugging this into our original equation gives:
$\lim\limits_{w \to \sqrt b} \frac{w - 2}{x} = \lim\limits_{w \to \sqrt b} \frac{w-2}{\frac{w^{2} - b}{a}}$
Multiply top and bottom by a
$\lim\limits_{w \to \sqrt b} \frac{w-2}{\frac{w^{2} - b}{a}} = \lim\limits_{w \to \sqrt b} \frac{a(w-2)}{w^{2}-b}$
$= \lim\limits_{w \to \sqrt b} \frac{a(w-2)}{(w-\sqrt b)(w + \sqrt b)}$
For this limit to exist we must let one of the terms in the denominator cancel with the term in the numerator. We choose the term $w - \sqrt b$, because the square root of something can never be negative. Therefore $\sqrt b = 2$
$ \lim\limits_{w \to \sqrt b} \frac{a(w-2)}{(w-\sqrt b)(w + \sqrt b)}$
$= \lim\limits_{w \to 2} \frac{a}{w+2}$
$= \frac{a}{4}$
$= 1$
Therefore $a = 4$. Since$\sqrt b = 2$, $b = 4$
