Đề bài
Câu 1: Tính \[\mathop {\lim }\limits_{x \to + \infty } \dfrac{5}{{3x + 2}}\]
A.0 B. 1
C. \[\dfrac{5}{3}\] D. \[ + \infty \]
Câu 2: Tìm giới hạn sau: \[\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - 3{x^2} + 2}}{{{x^2} - 4x + 3}}\]
A. \[ + \infty \] B. \[ - \infty \]
C. \[\dfrac{3}{2}\] D. 1
Câu 3: Cho hàm số \[f[x] = \dfrac{{x - 3}}{{\sqrt {{x^2} - 9} }}\]. Giá trị đúng của \[\mathop {\lim }\limits_{x \to {3^ + }} f[x]\] là
A.0 B. \[ - \infty \]
C. \[ + \infty \] D. \[\sqrt 6 \]
Câu 4: Tính giới hạn sau: \[\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt[3]{{1 + {x^4} + {x^6}}}}}{{\sqrt {1 + {x^3} + {x^4}} }}\]
A. \[ + \infty \] B. \[ - \infty \]
C. \[\dfrac{4}{3}\] D. 1
Câu 5: Tính \[\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}}\]
A.3 B. \[\dfrac{1}{2}\]
C. 1 D. \[ + \infty \]
Câu 6: Tìm giới hạn \[\mathop {\lim }\limits_{x \to - \infty } x[\sqrt {4{x^2} + 1} - x]\]
A. \[ + \infty \] B. \[ - \infty \]
C. \[\dfrac{4}{3}\] D. 0
Câu 7: Tính \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{[1 + 3x]}^3} - {{[1 - 4x]}^4}}}{x}\]
A. \[\dfrac{{ - 1}}{6}\] B. \[ - \infty \]
C. \[ + \infty \] D. 25
Câu 8: Tính \[\mathop {\lim }\limits_{x \to 0} \dfrac{{[1 + x][1 + 2x][1 + 3x] - 1}}{x}\]
A. \[ + \infty \] B. \[ - \infty \]
C. \[\dfrac{{ - 1}}{6}\] D. 6
Câu 9: Tính \[\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {2x + 3} - x}}{{{x^2} - 4x + 3}}\]
A.1 B. \[\dfrac{{ - 1}}{3}\]
C. \[ + \infty \] D. \[ - \infty \]
Câu 10: Tính \[\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x - \sqrt {3{x^2} + 2} }}{{5x + \sqrt {{x^2} + 1} }}\]
A. \[ + \infty \] B. \[ - \infty \]
C. \[\dfrac{{2 - \sqrt 3 }}{6}\] D. 0
Lời giải chi tiết
|
Câu |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Đáp án |
A |
C |
A |
D |
A |
B |
D |
D |
B |
C |
Câu 1: Đáp án A
\[\mathop {\lim }\limits_{x \to + \infty } \dfrac{5}{{3x + 2}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{5}{x}}}{{3 + \dfrac{2}{x}}} = \dfrac{0}{3} = 0\]
Câu 2: Đáp án C
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - 3{x^2} + 2}}{{{x^2} - 4x + 3}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{[x - 1][{x^2} - 2x - 2]}}{{[x - 1][x - 3]}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 2x - 2}}{{x - 3}}\\ = \dfrac{{{1^2} - 2.1 - 2}}{{1 - 3}} = \dfrac{3}{2}\end{array}\]
Câu 3: Đáp án A
\[\begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} f[x] = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{\sqrt {{x^2} - 9} }}\\ = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{\sqrt {[x - 3][x + 3]} }}\\ = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\sqrt {x - 3} }}{{\sqrt {x + 3} }} = \dfrac{0}{6} = 0\end{array}\]
Câu 4: Đáp án D
\[\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt[3]{{1 + {x^4} + {x^6}}}}}{{\sqrt {1 + {x^3} + {x^4}} }}\]
Câu 5: Đáp án A
\[\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\sqrt {{x^2} - x + 3} }}{{2x - 1}} = \dfrac{{\sqrt {{1^2} - 1 + 3} }}{{2.1 - 1}} = \sqrt 3 \]
Câu 6: Đáp án A
\[\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } x[\sqrt {4{x^2} + 1} - x]\\ = \mathop {\lim }\limits_{x \to - \infty } \left[ {\sqrt {4{x^4} + {x^2}} - {x^2}} \right]\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + {x^2}}}{{\sqrt {4{x^4} + {x^2}} + {x^2}}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^2} + 1}}{{\sqrt {4 + \dfrac{1}{{{x^2}}}} + 1}} = + \infty \end{array}\]
Câu 7: Đáp án D
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{[1 + 3x]}^3} - {{[1 - 4x]}^4}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 9x + 27{x^2} + 27{x^3} - {{\left[ {1 - 8x + 16{x^2}} \right]}^2}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 9x + 27{x^2} + 27{x^3} - 1 - 64{x^2} - 256{x^4} + 16x - 32{x^2} + 256{x^3}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{25x - 69{x^2} + 283{x^3} - 256{x^4}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left[ {25 - 69x + 283{x^2} - 256{x^3}} \right] = 25\end{array}\]
Câu 8: Đáp án D
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{[1 + x][1 + 2x][1 + 3x] - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {1 + 3x + 2{x^2}} \right]\left[ {1 + 3x} \right] - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{6x + 5{x^2} + 6{x^3}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left[ {6 + 5x + 6{x^2}} \right] = 6\end{array}\]
Câu 9: Đáp án B
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {2x + 3} - x}}{{{x^2} - 4x + 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{2x + 3 - {x^2}}}{{[x - 1][x - 3]\left[ {\sqrt {2x + 3} + x} \right]}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - [x - 3][x + 1]}}{{[x - 1][x - 3]\left[ {\sqrt {2x + 3} + x} \right]}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - [x + 1]}}{{[x - 1]\left[ {\sqrt {2x + 3} + x} \right]}}\\ = \dfrac{{ - [3 + 1]}}{{[3 - 1]\left[ {\sqrt {2.3 + 3} + 3} \right]}} = \dfrac{{ - 1}}{3}\end{array}\]
Câu 10: Đáp án C
\[\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x - \sqrt {3{x^2} + 2} }}{{5x + \sqrt {{x^2} + 1} }} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 - \sqrt {3 + \dfrac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \dfrac{1}{{{x^2}}}} }} = \dfrac{{2 - \sqrt 3 }}{6}\]