Question

Find the limit of the given function.
${lim}_{x\to 0}\frac{\sqrt{axb}-2}{x}=1$

Answer 1

(applying L' hospital Rule)
$={lim}_{x\to 0}\frac{\sqrt{axb}-2}{x}={lim}_{x\to 0}\frac{\frac{1}{2}(ax+b{)}^{\frac{-1}{2}}(a)}{1}=\frac{1}{2}(b{)}^{\frac{-1}{2}}a\frac{a}{2\sqrt{b}}$