Question

Find a &b so that

$\underset{x\to 0}{lt}\frac{\sqrt{ax+b}-3}{x}=\frac{1}{3}$

Answer 1

$\underset{x\to 0}{lim}\frac{\sqrt{ax+b}-3}{x}$

For limit to exist $\sqrt{b}=3⟹b=9$

Applying L-Hospital Rule

$\underset{x\to 0}{lim}\frac{\frac{a}{2\sqrt{ax+b}}}{1}=\frac{1}{3}$

$\frac{a}{6}=\frac{1}{3}⟹a=2$