mathop lim limitsx to 0 căn 2x + 1 - căn [3]x^2 + 1sin x
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\)
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - 1}}{{\sin x}} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{\sin x\left( {\sqrt {2x + 1} + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}}}{{\sin x\left( {1 + \sqrt[3]{{{x^2} + 1}} + {{\sqrt[3]{{{x^2} + 1}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}}\frac{2}{{\sqrt {2x + 1} + 1}} - \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}}.\frac{x}{{1 + \sqrt[3]{{{x^2} + 1}} + {{\sqrt[3]{{{x^2} + 1}}}^2}}}\\ = 1.\frac{2}{{1 + 1}} - 1.0 = 1\end{array}\)
Chọn A.
