Tìm các giới hạn sau:
Tìm các giới hạn sau:
Câu 1: \(\mathop {\lim }\limits_{x \to - 2} \dfrac{{{x^3} + 8}}{{\tan \left( {x + 2} \right)}}\)
A. \(12\)
B. \(4\)
C. \(8\)
D. \(2\)
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Đáp án : A(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - 2} \dfrac{{{x^3} + 8}}{{\tan \left( {x + 2} \right)}}\)
Đặt \(t = x + 2 \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 2} \dfrac{{{x^3} + 8}}{{\tan \left( {x + 2} \right)}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{{{\left( {t - 2} \right)}^3} + 8}}{{\tan t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\left( {t - 2 + 2} \right)\left[ {{{\left( {t - 2} \right)}^2} - 2\left( {t - 2} \right) + 4} \right]}}{{\tan t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{t\left[ {{{\left( {t - 2} \right)}^2} - 2\left( {t - 2} \right) + 4} \right]}}{{\tan t}}\\ = {\left( { - 2} \right)^2} - 2\left( { - 2} \right) + 4 = 12\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 2: \(\mathop {\lim }\limits_{x \to a} \dfrac{{{{\sin }^2}x - {{\sin }^2}a}}{{{x^2} - {a^2}}}\)
A. \(\dfrac{1}{{2a}}\sin 2a\)
B. \(\dfrac{1}{a}\sin a\)
C. \(\dfrac{1}{{3a}}\sin 2a\)
D. \(\dfrac{1}{2}\sin 2a\)
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Đáp án : A(2) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to a} \dfrac{{{{\sin }^2}x - {{\sin }^2}a}}{{{x^2} - {a^2}}}\)
Đặt \(t = x - a \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to a} \dfrac{{{{\sin }^2}x - {{\sin }^2}a}}{{{x^2} - {a^2}}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{{{\sin }^2}\left( {t + a} \right) - {{\sin }^2}a}}{{{{\left( {t + a} \right)}^2} - {a^2}}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\left[ {\sin \left( {t + a} \right) - \sin a} \right]\left[ {\sin \left( {t + a} \right) + \sin a} \right]}}{{\left( {t + a - a} \right)\left( {t + a + a} \right)}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{2\cos \left( {\dfrac{t}{2} + a} \right)\sin \dfrac{t}{2}.\left[ {\sin \left( {t + a} \right) + \sin a} \right]}}{{t\left( {t + a + a} \right)}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{2\cos \left( {\dfrac{t}{2} + a} \right).\dfrac{{\sin \dfrac{t}{2}}}{{\dfrac{t}{2}}}.\dfrac{t}{2}.\left[ {\sin \left( {t + a} \right) + \sin a} \right]}}{{t\left( {t + 2a} \right)}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\cos \left( {\dfrac{t}{2} + a} \right).\dfrac{{\sin \dfrac{t}{2}}}{{\dfrac{t}{2}}}.\left[ {\sin \left( {t + a} \right) + \sin a} \right]}}{{t + 2a}}\\ = \dfrac{{\cos a.1.\left( {\sin a + \sin a} \right)}}{{0+ 2a}} = \dfrac{{2\sin a\cos a}}{{2a}} = \dfrac{1}{{2a}}\sin 2a\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 3: \(\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin 2x}}{{1 + {{\cos }^3}x}}\)
A. \( - \infty \)
B. \(1\)
C. \( + \infty \)
D. \(0\)
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Đáp án : C(7) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin 2x}}{{1 + {{\cos }^3}x}}\)
Đặt \(t = x - \pi \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin 2x}}{{1 + {{\cos }^3}x}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sin 2\left( {t + \pi } \right)}}{{1 + {{\cos }^3}\left( {t + \pi } \right)}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sin 2t}}{{1 - \cos t}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{2\sin t\cos t}}{{2{{\sin }^2}\dfrac{t}{2}}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{2.2\sin \dfrac{t}{2}\cos \dfrac{t}{2}\cos t}}{{2{{\sin }^2}\dfrac{t}{2}}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{2\cos \dfrac{t}{2}\cos t}}{{\sin \dfrac{t}{2}}} = + \infty \end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 4: \(\mathop {\lim }\limits_{x \to b} \dfrac{{\cos x - \cos b}}{{x - b}}\)
A. \( - \sin b\)
B. \(\sin b\)
C. \( - \cos b\)
D. \(\cos b\)
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Đáp án : A(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to b} \dfrac{{\cos x - \cos b}}{{x - b}}\)
Đặt \(t = x - b \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to b} \dfrac{{\cos x - \cos b}}{{x - b}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{\cos \left( {t + b} \right) - \cos b}}{{t + b - b}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{ - 2\sin \left( {\dfrac{t}{2} + b} \right)\sin \dfrac{t}{2}}}{t}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{ - 2\sin \left( {\dfrac{t}{2} + b} \right)\dfrac{{\sin \dfrac{t}{2}}}{{\dfrac{t}{2}}}.\dfrac{t}{2}}}{t}\\ = \mathop {\lim }\limits_{t \to 0} \left[ { - \sin \left( {\dfrac{t}{2} + b} \right)\dfrac{{\sin \dfrac{t}{2}}}{{\dfrac{t}{2}}}} \right]\\ = - \sin b\end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
Câu 5: \(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}\)
A. \(0\)
B. \(1\)
C. \(\dfrac{1}{2}\)
D. \(\sqrt 2 \)
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Đáp án : D(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}\)
Đặt \(t = x - \dfrac{\pi }{4} \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - 2\cos \left( {t + \dfrac{\pi }{4}} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - 2\left( {\cos t.\dfrac{{\sqrt 2 }}{2} - \sin t.\dfrac{{\sqrt 2 }}{2}} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - \sqrt 2 \left( {\cos t - \sin t} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - \sqrt 2 \cos t + \sqrt 2 \sin t}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \left[ {\dfrac{{\sqrt 2 \left( {1 - \cos t} \right)}}{{\sin t}} + \sqrt 2 } \right]\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sqrt 2 .2{{\sin }^2}\dfrac{t}{2}}}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}} + \sqrt 2 } \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sqrt 2 \sin \dfrac{t}{2}}}{{\cos \dfrac{t}{2}}} + \sqrt 2 } \right)\\ = \dfrac{{\sqrt 2 .0}}{1} + \sqrt 2 = \sqrt 2 \end{array}\)
Lời giải sai Bình thường Khá hay Rất Hay
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