Tìm các giới hạn sau:
Tìm các giới hạn sau:
Câu 1: \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right)\)
A. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty
\end{array}\)
B. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty
\end{array}\)
C. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty
\end{array}\)
D. \(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty
\end{array}\)
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Đáp án : A(9) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right)\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + 3x + 1 - {x^2}}}{{\sqrt {{x^2} + 3x + 1} + x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3x + 1}}{{\sqrt {{x^2} + 3x + 1} + x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3 + \dfrac{1}{x}}}{{\sqrt {1 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}}} + 1}} = \dfrac{3}{2}\end{array}\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {1 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}}} - x} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left[ { - x\left( {\sqrt {1 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}}} + 1} \right)} \right] = + \infty \end{array}\)
Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - x} \right) = + \infty \\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {1 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}}} + 1} \right) = 2 > 0\end{array} \right.\).
Lời giải sai Bình thường Khá hay Rất Hay
Câu 2: \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {4{x^2} + 2x} + x - 1} \right)\)
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 \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\\mathop {\lim }\limits_{x \to + \infty } \left( { - \sqrt {4 + \dfrac{2}{x}} + 1 - \dfrac{1}{x}} \right) = - 1 < 0\end{array} \right.\).
Lời giải sai Bình thường Khá hay Rất Hay
Câu 3: \(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\)
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:
\(\mathop {\lim }\limits_{x \to \pm \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } x.\dfrac{{{x^2} + 1 - {x^2} + 3}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{4}{{\sqrt {1 + \dfrac{1}{{{x^2}}}} + \sqrt {1 - \dfrac{3}{{{x^2}}}} }} = 2\end{array}\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 3} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{{x^2} + 1 - {x^2} + 3}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - 3} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{4}{{ - \sqrt {1 + \dfrac{1}{{{x^2}}}} - \sqrt {1 - \dfrac{3}{{{x^2}}}} }} = - 2\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 4: \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\)
A. \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right) = \dfrac{1}{2}\) \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\) không tồn tại.
B. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right) = \dfrac{1}{2}\)
C. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right) = -\dfrac{1}{2}\)
D. \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\) không tồn tại
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Đáp án : A(4) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3x + \sqrt {3x + \sqrt {3x} } - 3x}}{{\sqrt {3x + \sqrt {3x + \sqrt {3x} } } + \sqrt {3x} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {3x + \sqrt {3x} } }}{{\sqrt {3x + \sqrt {3x + \sqrt {3x} } } + \sqrt {3x} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {3 + \dfrac{{\sqrt {3x} }}{x}} }}{{\sqrt {3 + \dfrac{{\sqrt {3x + \sqrt {3x} } }}{x}} + \sqrt 3 }}\\ = \dfrac{{\sqrt 3 }}{{\sqrt 3 + \sqrt 3 }} = \dfrac{1}{2}\end{array}\)
\(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } } - \sqrt {3x} } \right)\) không tồn tại do \(\sqrt {3x} \) xác định khi và chỉ khi \(3x \ge 0 \Leftrightarrow x \ge 0\).
Lời giải sai Bình thường Khá hay Rất Hay
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