Đề bài
Câu 1: Giá trị đúng của lim
A.Không tồn tại B. 0
C. 1 D. + \infty
Câu 2: Tìm giới hạn D = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{x + 1}} - 1}}{{\sqrt {2x + 1} - 1}}:
A. + \infty B. - \infty
C. \dfrac{1}{3} D. 0
Câu 3: Tìm giới hạn A = \mathop {\lim }\limits_{x \to 2} \dfrac{{2{x^2} - 5x + 2}}{{{x^3} - 3x - 2}}:
A. + \infty B. - \infty
C. \dfrac{1}{3} D. 1
Câu 4: \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} - x + 1}}{{{x^2} - 1}} bằng:
A. - \infty B. -1
C. 1 D. + \infty
Câu 5: Tìm giới hạn E = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} - x} \right):
A. + \infty B. - \infty
C. - \dfrac{1}{2} D. 0
Câu 6: Tìm giới hạn E = \mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {4{x^2} + 1} - x} \right):
A. + \infty B. - \infty
C. \dfrac{4}{3} D. 0
Câu 7: Chọn kết quả đúng của \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{1}{{{x^2}}} - \dfrac{2}{{{x^3}}}} \right):
A. - \infty B. 0
C. + \infty D. Không tồn tại
Câu 8: Tìm giới hạn B = \mathop {\lim }\limits_{x \to - \infty } \left( {x - \sqrt {{x^2} + x + 1} } \right):
A. + \infty B. - \infty
C. \dfrac{4}{3} D. 0
Câu 9: Tìm giới hạn A = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {4x + 1} - \sqrt[3]{{2x + 1}}}}{x}:
A. + \infty B. - \infty
C. 2 D. 0
Câu 10: Tìm giới hạn C = \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt[{}]{{2x + 3}} - 3}}{{{x^2} - 4x + 3}}:
A. + \infty B. - \infty
C. \dfrac{1}{6} D. 0
Lời giải chi tiết
| Câu | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Đáp án | A | C | C | D | C | B | C | B | C | C |
Câu 1: Đáp án A
TXD: R/\left\{ 3 \right\}
\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{x - 3}} = 1, (x-3>0, với mọi x>3)
\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{\left| {x - 3} \right|}}{{x - 3}} = - 1, (x-3<0, với mọi x<3)
\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{x - 3}} \ne \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{\left| {x - 3} \right|}}{{x - 3}}, nên không tồn tại giới hạn khi x \to 3
Câu 2: Đáp án C
\begin{array}{l}D = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{x + 1}} - 1}}{{\sqrt {2x + 1} - 1}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt[3]{{x + 1}} - 1} \right)\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)}}{{\left( {\sqrt {2x + 1} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {\sqrt {2x + 1} + 1} \right)}}{{2x\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {2x + 1} + 1} \right)}}{{2\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} - 1} \right)}}\\ = \dfrac{2}{{2(1 + 1 + 1)}} = \dfrac{1}{3}\end{array}
Câu 3: Đáp án C
\begin{array}{l}A = \mathop {\lim }\limits_{x \to 2} \dfrac{{2{x^2} - 5x + 2}}{{{x^3} - 3x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2x - 1} \right)\left( {x - 2} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2x - 1}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{3}{9} = \dfrac{1}{3}\end{array}
Câu 4: Đáp án D
TXD R/\left\{ 1 \right\}
\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - x + 1} \right) = 1 \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - 1} \right) = 0và \left( {{x^2} - 1} \right) > 0
\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} - x + 1}}{{{x^2} - 1}} = + \infty
Câu 5: Đáp án C
\begin{array}{l}E = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} - x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - x + 1}}{{\left( {\sqrt {{x^2} - x + 1} + x} \right)}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 1 + \dfrac{1}{x}}}{{\left( {\sqrt {1 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + 1} \right)}} = \dfrac{{ - 1}}{2}\end{array}
Câu 6: Đáp án B
\begin{array}{l}E = \mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {4{x^2} + 1} - x} \right) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {3{x^2} + 1} \right)}}{{\left( {\sqrt {4{x^2} + 1} + x} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^3}\left( {3 + \dfrac{1}{{{x^2}}}} \right)}}{{{x^3}\left( {\sqrt {\dfrac{4}{{{x^2}}} + \dfrac{1}{{{x^6}}}} + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3 + \dfrac{1}{{{x^2}}}} \right)}}{{\left( {\sqrt {\dfrac{4}{{{x^2}}} + \dfrac{1}{{{x^6}}}} + \dfrac{1}{{{x^2}}}} \right)}} = - \infty \end{array}
Câu 7: Đáp án C
\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{1}{{{x^2}}} = + \infty \\\mathop {\lim }\limits_{x \to {0^ - }} \left( { - \dfrac{2}{{{x^3}}}} \right) = + \infty \\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{1}{{{x^2}}} - \dfrac{2}{{{x^3}}}} \right) = + \infty \end{array}
Câu 8: Đáp án B
\begin{array}{l}B = \mathop {\lim }\limits_{x \to - \infty } \left( {x - \sqrt {{x^2} + x + 1} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left( {x - \left| x \right|\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } x\left( {1 + \sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} } \right) = - \infty \end{array}
Câu 9: Đáp án C
\begin{array}{l}A = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {4x + 1} - \sqrt[3]{{2x + 1}}}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4x + 1 - \sqrt[3]{{{{\left( {2x + 1} \right)}^2}}}}}{{x\left( {\sqrt {4x + 1} + \sqrt[3]{{2x + 1}}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4 + \dfrac{1}{x} - \sqrt[3]{{\dfrac{2}{x} + \dfrac{2}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}{{\left( {\sqrt {4x + 1} + \sqrt[3]{{2x + 1}}} \right)}} = \dfrac{4}{2} = 2\end{array}
Câu 10: Đáp án C
\begin{array}{l}C = \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt[{}]{{2x + 3}} - 3}}{{{x^2} - 4x + 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{2(x - 3)}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 3} + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{2}{{\left( {x - 1} \right)\left( {\sqrt {2x + 3} + 3} \right)}} = \dfrac{1}{6}\end{array}
