Answer
$a=b=4$
Work Step by Step
$$\lim_{x \to 0}{\frac{(\sqrt {ax+b})^{2}-2^{2}}{x(\sqrt {ax+b}+2)}}$$
$$\lim_{x \to 0}{\frac{ax+b-4}{x(\sqrt {ax+b}+2)}}$$
So to cancel out $x$ from the denominator and the numerator of the fraction, $b-4$ should be equal to $0$:
$$b-4=0\to b=4$$
$$\lim_{x \to 0}{\frac{ax+4-4}{x(\sqrt {ax+4}+2)}}$$
$$\lim_{x \to 0}{\frac{a}{\sqrt {ax+4}+2}}=1$$
$${\frac{a}{\sqrt {a\cdot 0+4}+2}}=1$$
$${\frac{a}{\sqrt {4}+2}}=1$$
$${\frac{a}{4}}=1$$
$$a=4$$
