Đề bài

Câu 1: Giá trị đúng của \(\mathop {\lim }\limits_{x \to 3} \dfrac{{\left| {x - 3} \right|}}{{x - 3}}\)

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 \(R/\left\{ 1 \right\}\)

\(\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - x + 1} \right) = 1\)                              \(\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - 1} \right) = 0\)và \(\left( {{x^2} - 1} \right) > 0\)

\(\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} - x + 1}}{{{x^2} - 1}} =  + \infty \)

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

\(\begin{array}{l}C = \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt[{}]{{2x + 3}} - 3}}{{{x^2} - 4x + 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{2(x - 3)}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 3}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{2}{{\left( {x - 1} \right)\left( {\sqrt {2x + 3}  + 3} \right)}} = \dfrac{1}{6}\end{array}\)

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