Đáp án:
23. B
Giải thích các bước giải:
\(Đề= \lim\limits_{x\to0} \dfrac{\sqrt{1+ax}.\sqrt[3]{1+bx}-\sqrt{1+ax}+\sqrt{1+ax}-1}{x}\)
=\(\lim\limits_{x\to0} \dfrac{\sqrt{1+ax}.(\sqrt[3]{1+bx}-1)}{x}+\lim\limits_{x\to0} \dfrac{\sqrt{1+ax}-1}{x}\)
=\(\lim\limits_{x\to0} \dfrac{bx.\sqrt{1+ax}}{x.[(\sqrt[3]{1+bx})^{2}+\sqrt[3]{1+bx}+1]}+\lim\limits_{x\to0} \dfrac{ax}{x\sqrt{1+ax}+1}\)
=\(\lim\limits_{x\to0} \dfrac{b.\sqrt{1+ax}}{[(\sqrt[3]{1+bx})^{2}+\sqrt[3]{1+bx}+1]}+\lim\limits_{x\to0} \dfrac{a}{\sqrt{1+ax}+1}\)
=\(\dfrac{b}{3}+\dfrac{a}{2}\)
