\(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\)
Tìm các giới hạn sau:
Câu 475606: \(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\)
A. \(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) =- 2\)
B. \(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) = 2\)
C. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) = -2\\
\mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) = 2
\end{array}\)
D. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) = 2\\
\mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right) = - 2
\end{array}\)
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Đáp án : D(15) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } x.\dfrac{{{x^2} + 1 - {x^2} + 3}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{4}{{\sqrt {1 + \dfrac{1}{{{x^2}}}} + \sqrt {1 - \dfrac{3}{{{x^2}}}} }} = 2\end{array}\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{{x^2} + 1 - {x^2} + 3}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{4}{{ - \sqrt {1 + \dfrac{1}{{{x^2}}}} - \sqrt {1 - \dfrac{3}{{{x^2}}}} }} = - 2\end{array}\)
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