Tính giới hạn \(B=\lim\limits _{x \rightarrow+\infty} \frac{x \sqrt{x^{2}+1}-2 x+1}{\sqrt[3]{2 x^{3}-2}+1}\).
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Lời giải:
Báo sai\(\begin{array}{l} B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}}} \right)}}{{x\left( {\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x}} \right)}} = \frac{{x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}}} \right)}}{{\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x}}} = + \infty \\ \left(Vì {\left\{ \begin{array}{l} \mathop {\lim }\limits_{x \to + \infty } = + \infty \\ \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}}}}{{\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x}}} = \frac{1}{{\sqrt[3]{2}}} > 0 \end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}}} \right)}}{{\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x}}} = + \infty } \right) \end{array}\)