\(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\)
Tìm các giới hạn sau:
Câu 475605: \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\)
A. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x + x} - 1} \right) = 0\)
B. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x + x} - 1} \right) = - \infty \)
C. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x + x} - 1} \right) = + \infty \)
D. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x + x} - 1} \right) = \pm \infty \)
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Đáp án : C(10) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\)
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {x\sqrt {4 + \dfrac{2}{x}} + x - 1} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {4 + \dfrac{2}{x}} + 1 - \dfrac{1}{x}} \right) = + \infty \end{array}\)
Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } x = + \infty \\\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {4 + \dfrac{2}{x}} + 1 - \dfrac{1}{x}} \right) = 3 > 0\end{array} \right.\).
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {4 + \dfrac{2}{x}} + x - 1} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } x\left( { - \sqrt {4 + \dfrac{2}{x}} + 1 - \dfrac{1}{x}} \right) = + \infty \end{array}\)
Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\\mathop {\lim }\limits_{x \to + \infty } \left( { - \sqrt {4 + \dfrac{2}{x}} + 1 - \dfrac{1}{x}} \right) = - 1 < 0\end{array} \right.\).
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