Question Number 161002 by bobhans last updated on 10/Dec/21
$$\:\:\:\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ax}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{2x}}\:\sqrt{\mathrm{1}+\mathrm{bx}}\:−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{then}\:\mathrm{a}×\mathrm{b}\:=? \\ $$
Answered by blackmamba last updated on 11/Dec/21
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ax}}{\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{{bx}}{\mathrm{2}}\right)−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ax}}{\left(\frac{\mathrm{2}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}\right){x}+\frac{\mathrm{2}{bx}^{\mathrm{2}} }{\mathrm{6}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\frac{\mathrm{6}{a}}{\mathrm{4}+\mathrm{3}{b}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}=\frac{\mathrm{4}+\mathrm{3}{b}}{\mathrm{12}} \\ $$$$\:\:{a}×{b}=\:\frac{\mathrm{4}{b}+\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{12}} \\ $$