How to find $a$ and $b$ in $\lim\limits_{x→0}\frac{\sqrt{ax+b}-2}{x}=1$? How to find $a$ and $b$ such that $$\lim_{x→0}\frac{\sqrt{ax+b}-2}{x}=1?$$
By setting the equation equal to one directly I was able to find out that $b = 4$. However, when I try to solve for $a$ I am stuck and left in circles. How would I go about solving this problem?

You could also do it using binomial expansion or Taylor series (built at $x=0$)
$$\sqrt{a x+b}=\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right)$$ making 
$$\frac{\sqrt{ax+b}-2}{x}=\frac{-2+\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right) }x=\frac{\sqrt{b}-2}{x}+\frac{a}{2 \sqrt{b}}-\frac{a^2 x}{8 b^{3/2}}+O\left(x^2\right)$$ So
$$\frac{a}{2 \sqrt{b}}=1 \qquad \text{and}\qquad \sqrt{b}-2=0$$ then $a,b$.
