Cho a,b,c là các số thực khác 0, 3b-2c khác 0. Tìm hệ thức liên hệ giữa a,b,c để lim [{tan(ax)} / {căn(1+bx) - căn3(1+cx)}] = 1/2 x-->0 a. a / (3b-2c) = 1/10 b. a / (3b-2c) = 1/6 c. a / (3b-2c) = 1/2 d. a / (3b-2c) = 1/12

Đáp án: $D$

Giải thích các bước giải:

Ta có:

$\lim_{x\to0}\dfrac{\tan(ax)}{\sqrt{1+bx}-\sqrt[3]{1+cx}}=\dfrac12$

$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{(\sqrt{1+bx}-1)+(1-\sqrt[3]{1+cx})}=\dfrac12$

$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{\dfrac{1+bx-1}{\sqrt{1+bx}+1}+\dfrac{1-(1+cx)}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$

$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{\cos(ax)}}{\dfrac{bx}{\sqrt{1+bx}+1}+\dfrac{-cx}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$

$\to \lim_{x\to0}\dfrac{\dfrac{\sin(ax)}{ax}\cdot \dfrac{a}{\cos(ax)}}{\dfrac{b}{\sqrt{1+bx}+1}-\dfrac{c}{1+\sqrt[3]{1+cx}+(\sqrt[3]{1+cx})^2}}=\dfrac12$

$\to \dfrac{1\cdot \dfrac{a}{\cos(a\cdot 0)}}{\dfrac{b}{\sqrt{1+b\cdot0}+1}-\dfrac{c}{1+\sqrt[3]{1+c\cdot0}+(\sqrt[3]{1+c\cdot0})^2}}=\dfrac12$

$\to \dfrac{6a}{3b-2c}=\dfrac12$

$\to \dfrac{a}{3b-2c}=\dfrac1{12}$

$\to D$