Tìm các giới hạn sau:
Tìm các giới hạn sau:
Câu 1: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\)
A. \(0\)
B. \(1\)
C. \(-1\)
D. \(\dfrac{1}{2}\)
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Đáp án : B(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - 1 + 1 - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - 1}}{{\sin x}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x + 1 - 1}}{{\sin x\left( {\sqrt {2x + 1} + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - {x^2} - 1}}{{\sin x\left( {1 + {x^2} + 1 + {{\sqrt[3]{{{x^2} + 1}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{\sin x}}.\dfrac{2}{{\sqrt {2x + 1} + 1}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{\sin x}}.\dfrac{x}{{1 + \sqrt[3]{{{x^2} + 1}} + {{\sqrt[3]{{{x^2} + 1}}}^2}}}\\ = 1.\dfrac{2}{{1 + 1}} - 1.\dfrac{0}{{1 + 1 + {1^2}}} = 1\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 2: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + \tan x} - \sqrt {1 + \sin x} }}{{{x^3}}}\)
A. \(\dfrac{1}{4}\)
B. \(\dfrac{1}{2}\)
C. \(1\)
D. \(2\)
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Đáp án : A(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + \tan x} - \sqrt {1 + \sin x} }}{{{x^3}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + \tan x - 1 - \sin x}}{{{x^3}\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin x}}{{\cos x}} - \sin x}}{{{x^3}\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\left( {1 - \cos x} \right)}}{{{x^3}\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x.2{{\sin }^2}\dfrac{x}{2}}}{{{x^3}\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2.\dfrac{{\sin x}}{x}.x.{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}}}{{{x^3}\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{2}.\dfrac{{\sin x}}{x}.{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}}}{{\left( {\sqrt {1 + \tan x} + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \dfrac{{\dfrac{1}{2}.1.1}}{{\left( {1 + 1} \right).1}} = \dfrac{1}{4}\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 3: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - \cos x - x}}{{{x^2}}}\)
A. \(1\)
B. \(-1\)
C. \(0\)
D. \(\dfrac{1}{2}\)
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Đáp án : C(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - \cos x - x}}{{{x^2}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - \cos x - x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - x - 1 + 1 - \cos x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - \left( {x + 1} \right)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 2x - {{\left( {x + 1} \right)}^2}}}{{{x^2}\left[ {\sqrt {1 + 2x} + \left( {x + 1} \right)} \right]}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{\sqrt {1 + 2x} + \left( {x + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}}}{{{x^2}}}\\ = \dfrac{{ - 1}}{{\sqrt 1 + 1}} + \mathop {\lim }\limits_{x \to 0} \dfrac{1}{2}{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2}\\ = \dfrac{{ - 1}}{2} + \dfrac{1}{2} = 0\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 4: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
A. \(\dfrac{1}{3}\)
B. \(\dfrac{1}{2}\)
C. \(\dfrac{1}{6}\)
D. \(\dfrac{1}{4}\)
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Đáp án : C(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \sqrt[3]{{\cos x}}} \right)\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}{{{{\tan }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos x} \right){{\cos }^2}x}}{{{{\sin }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}{{\cos }^2}x}}{{{{\sin }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}{{\cos }^2}x}}{{{{\left( {\dfrac{{\sin x}}{x}} \right)}^2}.{x^2}\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{2}{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.{{\cos }^2}x}}{{{{\left( {\dfrac{{\sin x}}{x}} \right)}^2}.\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \dfrac{{\dfrac{1}{2}{1^2}.1}}{{{1^2}.\left( {1 + 1 + {1^2}} \right)}} = \dfrac{1}{6}\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
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