Đề kiểm tra 15 phút - Đề số 5 - Chương 4 - Đại số và Giải tích 11

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: 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

\(\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x - \sqrt {3{x^2} + 2} }}{{5x + \sqrt {{x^2} + 1} }} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2 - \sqrt {3 + \dfrac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \dfrac{1}{{{x^2}}}} }} = \dfrac{{2 - \sqrt 3 }}{6}\)