Tìm giới hạn \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{x + 1}} - 1}}{{\sqrt[4]{{2x + 1}} - 1}}\)

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{x + 1}} - 1}}{{\sqrt[4]{{2x + 1}} - 1}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{(\sqrt[3]{{x + 1}} - 1)(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)(\sqrt[4]{{2x + 1}} + 1)}}{{(\sqrt[4]{{2x + 1}} - 1)(\sqrt[4]{{2x + 1}} + 1)(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x(\sqrt[4]{{2x + 1}} + 1)}}{{(\sqrt[2]{{2x + 1}} - 1)(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x(\sqrt[4]{{2x + 1}} + 1)(\sqrt[2]{{2x + 1}} + 1)}}{{(\sqrt[2]{{2x + 1}} - 1)(\sqrt[2]{{2x + 1}} + 1)(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x(\sqrt[4]{{2x + 1}} + 1)(\sqrt[2]{{2x + 1}} + 1)}}{{2x(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{(\sqrt[4]{{2x + 1}} + 1)(\sqrt[2]{{2x + 1}} + 1)}}{{2(\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2.2}}{{2(1 + 1 + 1)}} = \dfrac{2}{3}\end{array}\)