Tính \(B=\lim\limits _{x \rightarrow 0} \frac{\sqrt{1+2 x}-\sqrt[3]{1+3 x}}{1-\sqrt{\cos 2 x}}\).
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Lời giải:
Báo saiTa có:
\(B=\lim\limits _{x \rightarrow 0} \frac{\frac{\sqrt{1+2 x}-\sqrt[3]{1+3 x}}{x^{2}}}{\frac{1-\sqrt{\cos 2 x}}{x^{2}}}\)
\(\begin{array}{*{20}{l}} {{\rm{ Mà : }}\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - (1 + x)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \frac{{(x + 1) - \sqrt[3]{{1 + 3x}}}}{{{x^2}}}}\\ { = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{\sqrt {1 + 2x} + x + 1}} + \mathop {\lim }\limits_{x \to 0} \frac{{x + 3}}{{{{(x + 1)}^2} + (x + 1)\sqrt[3]{{1 + 3x}} + \sqrt[3]{{{{(1 + 3x)}^2}}}}}}\\ { = - \frac{1}{2} + 1 = \frac{1}{2}.}\\ {\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt {\cos 2x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{{x^2}}} \cdot \frac{1}{{1 + \sqrt {\cos 2x} }} = 1} \end{array}\)
\(\text { Vậy } \mathrm{B}=\frac{1}{2} \text { . }\)