Đáp án: $D$
Giải thích các bước giải:
Ta có:
$\lim_{x\to0}\dfrac{\tan(ax)}{\sqrt{1+bx}-\sqrt[3]{1+cx}}=\dfrac12$
$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{(\sqrt{1+bx}-1)+(1-\sqrt[3]{1+cx})}=\dfrac12$
$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{\dfrac{1+bx-1}{\sqrt{1+bx}+1}+\dfrac{1-(1+cx)}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$
$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{\dfrac{bx}{\sqrt{1+bx}+1}+\dfrac{-cx}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$
$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{ax}\cdot \dfrac{a}{\cos(ax)}}{\dfrac{b}{\sqrt{1+bx}+1}-\dfrac{c}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$
$\to \dfrac{1\cdot \dfrac{a}{\cos(a\cdot 0)}}{\dfrac{b}{\sqrt{1+b\cdot0}+1}-\dfrac{c}{1+\sqrt[3]{1+c\cdot0}+(\sqrt[3]{1+c\cdot0})^2}}=\dfrac12$
$\to \dfrac{6a}{3b-2c}=\dfrac12$
$\to \dfrac{a}{3b-2c}=\dfrac1{12}$
$\to D$
