Đề bài
Câu 1: Tính \(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^3} + 2{x^2} + 1}}{{2{x^5} + 1}}\)
A.-2 B. \(\dfrac{{ - 1}}{2}\)
C. \(\dfrac{1}{2}\) \(D. 2\)
Câu 2: Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\)
A. \( - \infty \) B. 0
C. \(\dfrac{1}{2}\) D. \( + \infty \)
Câu 3: Tìm \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} - 2}}{{x - 1}}\)
A. \( + \infty \) B. \( - \infty \)
C. -2 D. \(\dfrac{1}{4}\)
Câu 4: Tính \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x + 3}}{{x - 2}}\)
A. \( + \infty \) B. \( - \infty \)
C. 1 D. -2
Câu 5: Tìm a để hàm số \(f(x) = \left\{ {\begin{array}{*{20}{c}}{{x^2} + ax + 1}\\{2{x^2} - x + 1\,}\end{array}} \right.\,\,\,\begin{array}{*{20}{c}}{khi}\\{khi}\end{array}\,\,\,\begin{array}{*{20}{c}}{\,\,x > 2}\\{x \le 2}\end{array}\) có giới hạn khi \(x \to 2\)
A. \(\dfrac{1}{2}\) B. \( + \infty \)
C. \( - \infty \) \(D. 1\)
Câu 6: Cho hàm số \(f(x) = \sqrt {\dfrac{{4{x^2} - 3x}}{{(2x - 1)({x^3} - 2)}}} \). Chọn kết quả đúng của \(\mathop {\lim }\limits_{x \to 2} f(x)\)
A. \(\dfrac{5}{9}\) B. \(\dfrac{{\sqrt 5 }}{3}\)
C. \(\dfrac{{\sqrt 5 }}{9}\) D. \(\dfrac{{\sqrt 2 }}{9}\)
Câu 7: Tính \(\mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{3x - 1}}{{x - 2}}\)
A. \( + \infty \) B. -2
C. 1 D. \( - \infty \)
Câu 8: Cho hàm số \(f(x) = \left\{ {\begin{array}{*{20}{c}}{{x^2} - 3}\\{x - 1\,}\end{array}} \right.\,\,\,\begin{array}{*{20}{c}}{khi}\\{khi}\end{array}\,\,\,\begin{array}{*{20}{c}}{\,\,x \ge 2}\\{x < 2}\end{array}\). Chọn kết quả đúng của \(\mathop {\lim }\limits_{x \to 2} f(x)\)
A.-1 B. 0
C. 1 D. Không tồn tại
Câu 9: Tính \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} - x + 1}}{{x + 2}}\)
A. \( + \infty \) B. \( - \infty \)
C. 1 D. -2
Câu 10: Tìm a để hàm số \(f(x) = \left\{ {\begin{array}{*{20}{c}}{5a{x^2} + 3x + 2a + 1}\\{1 + x + \sqrt {{x^2} + x + 2} \,}\end{array}} \right.\,\,\,\begin{array}{*{20}{c}}{khi}\\{khi}\end{array}\,\,\,\begin{array}{*{20}{c}}{\,\,x \ge 0}\\{x < 0}\end{array}\) có giới hạn khi\(x \to 0\)
A. \( + \infty \) B. \( - \infty \)
C. \(\dfrac{{\sqrt 2 }}{2}\) D. 1
Lời giải chi tiết
| Câu | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Đáp án | A | B | D | C | A | B | D | C | B | C |
Câu 1 : Đáp án A
\(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^3} + 2{x^2} + 1}}{{2{x^5} + 1}} = \dfrac{{{{\left( { - 1} \right)}^3} + 2.{{\left( { - 1} \right)}^2} + 1}}{{2.{{\left( { - 1} \right)}^5} + 1}} = - 2\)
Câu 2 : Đáp án B
\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\\ = \mathop {\lim }\limits_{x \to - 1} \dfrac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to - 1} \dfrac{{\left( {x + 1} \right)}}{{2\left( {{x^2} - x + 1} \right)}}\\ = \dfrac{{ - 1 + 1}}{{2\left( {1 - 1 + 1} \right)}} = 0\end{array}\)
Câu 3 : Đáp án D
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} - 2}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {x + 3} - 2} \right)\left( {\sqrt {x + 3} + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {\sqrt {x + 3} + 2} \right)}} = \dfrac{1}{4}\end{array}\)
Câu 4 : Đáp án C
\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x + 3}}{{x - 2}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {1 + \dfrac{3}{x}} \right)}}{{x\left( {1 - \dfrac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {1 + \dfrac{3}{x}} \right)}}{{\left( {1 - \dfrac{2}{x}} \right)}} = 1\)
Câu 5 : Đáp án D
\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} {x^2}{\rm{ + ax + 1 = 5 + 2a}}\) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} 2{x^2}{\rm{ - x + 1 = 7}}\)
Để hàm số có giới hạn khi \(x \to 2\) thì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \Leftrightarrow 5 + 2a = 7 \Leftrightarrow a = 1\)
Câu 6 : Đáp án B
\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \sqrt {\dfrac{{4{x^2} - 3x}}{{(2x - 1)({x^3} - 2)}}} = \sqrt {\dfrac{{{{4.2}^2} - 3.2}}{{\left( {2.2 - 1} \right)\left( {{2^3} - 2} \right)}}} = \dfrac{{\sqrt 5 }}{3}\)
Câu 7 : Đáp án D
Hàm số f(x) xác định trên \(R/\left\{ 2 \right\}\)
Ta có \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 0\), x-2<0, với mọi x<2 , \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {3x - 1} \right) = 5 > 0\)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{3x - 1}}{{x - 2}} = - \infty \)
Câu 8 : Đáp án C
\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} {x^2} - 3 = 1\) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} x - 1 = 1\)
\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 1\)
Câu 9 : Đáp án B
\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} - x + 1}}{{x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2}\left( {2 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)}}{{{x^2}\left( {\dfrac{1}{x} + \dfrac{2}{{{x^2}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {2 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)}}{{\left( {\dfrac{1}{x} + \dfrac{2}{{{x^2}}}} \right)}} = - \infty \\\mathop {\lim }\limits_{x \to - \infty } \left( {2 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right) = 2\\\mathop {\lim }\limits_{x \to - \infty } \left( {\dfrac{1}{x} + \dfrac{2}{{{x^2}}}} \right) = - \infty \end{array}\)
Câu 10 : Đáp án C
\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {5a{x^2} + 3x + 2a + 1} \right) = 2a + 1\)
\(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {1 + x + \sqrt {{x^2} + x + 2} } \right) = 1 + \sqrt 2 \)
Để hàm số có giới hạn khi\(x \to 0\) thì \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \Leftrightarrow 2a + 1 = 1 + \sqrt 2 \Rightarrow a = \dfrac{1}{{\sqrt 2 }}\)
