29 câu trắc nghiệm Toán 11 Chân trời sáng tạo Giới hạn của hàm số có đáp án

Vì giới hạn đã cho tồn tại nên \[\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \left( {\sqrt {\left( {{\rm{a + 5}}} \right){{\rm{x}}^{\rm{2}}} - {\rm{2}}\left( {{\rm{a + 2}}} \right){\rm{x + 2a + b + 7}}} - \sqrt {{\rm{6x + 3}}} } \right){\rm{ = 0}}\]

\[ \Rightarrow \sqrt {{\rm{a + b + 8}}} - {\rm{3 = 0}} \Rightarrow {\rm{b = 1}} - {\rm{a}}\]

Khi đó\[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {\left( {{\rm{a + 5}}} \right){{\rm{x}}^{\rm{2}}} - {\rm{2}}\left( {{\rm{a + 2}}} \right){\rm{x + 2a + b + 7}}} - \sqrt {{\rm{6x + 3}}} }}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x + 1}}}}{\rm{ = }}\frac{{13}}{{12}}\]

\[ \Rightarrow \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{\sqrt {\left( {{\rm{a + 5}}} \right){{\rm{x}}^{\rm{2}}} - {\rm{2}}\left( {{\rm{a + 2}}} \right){\rm{x + a + 8}}} - \sqrt {{\rm{6x + 3}}} }}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x + 1}}}}{\rm{ = }}\frac{{{\rm{13}}}}{{{\rm{12}}}}\]

\[ \Rightarrow \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{{\rm{a + 5}}}}{{\left( {\sqrt {{\rm{(a + 5)}}{{\rm{x}}^{\rm{2}}} - {\rm{2(a + 2)x + a + 8}}} {\rm{ + }}\sqrt {{\rm{6x + 3}}} } \right)}}{\rm{ = }}\frac{{{\rm{13}}}}{{{\rm{12}}}}\]

\[ \Leftrightarrow \frac{{{\rm{a}} + 5}}{6} = \frac{{13}}{{12}} \Leftrightarrow {\rm{a}} = \frac{3}{2} \Rightarrow {\rm{b}} = - \frac{1}{2} \Rightarrow {{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}} = \frac{5}{2}\]

Chọn đáp án B

Đáp án cần chọn là: B

Vì\[\mathop {\lim }\limits_{{\rm{x}} \to 3} \frac{{{\rm{P}}\left( {\rm{x}} \right) - 2}}{{{\rm{x}} - 3}} = 2 \Rightarrow {\rm{P}}\left( 3 \right) - 2 = 0 \Rightarrow {\rm{P}}\left( {\rm{3}} \right) = 2\]

Ta có:\[\mathop {\lim }\limits_{{\rm{x}} \to 3} \frac{{{\rm{P}}\left( {\rm{x}} \right) - 2}}{{\left( {{{\rm{x}}^2} - 9} \right)\left( {\sqrt {{\rm{P}}\left( {\rm{x}} \right) + 2} + 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 3} \frac{{{\rm{P}}\left( {\rm{x}} \right) - 2}}{{\left( {{\rm{x}} - 3} \right)}}.\frac{1}{{\left( {{\rm{x}} + 3} \right)\left( {\sqrt {{\rm{P}}\left( {\rm{x}} \right) + 2} + 1} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 3} \frac{{{\rm{P}}\left( {\rm{x}} \right) - 2}}{{\left( {{\rm{x}} - 3} \right)}}.\mathop {\lim }\limits_{{\rm{x}} \to 3} \frac{1}{{\left( {{\rm{x}} + 3} \right)\left( {\sqrt {{\rm{P}}\left( {\rm{x}} \right) + 2} + 1} \right)}} = 2.\frac{1}{{\left( {3 + 3} \right)\left( {\sqrt {2 + 2} + 1} \right)}} = \frac{1}{9}\]

Chọn đáp án C

Đáp án cần chọn là: C

Ta có\[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {2020{{\rm{x}}^2} + {\rm{x}} + 3} - \sqrt {2021{{\rm{x}}^2} + 2} } \right) = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left[ {{\rm{x}}\left( {\sqrt {2020 + \frac{1}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right)} \right]\]

Vì\[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } {\rm{x}} = + \infty ,\mathop {\lim }\limits_{{\rm{x}} \to \infty } \left( {\sqrt {2020 + \frac{1}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right) = \sqrt {2020} - \sqrt {2021} < 0\]

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left[ {{\rm{x}}\left( {\sqrt {2020 + \frac{{\rm{1}}}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right)} \right] = - \infty \]

Chọn đáp án A

Đáp án cần chọn là: A

Vì\[\mathop {\lim }\limits_{{\rm{x}} \to 4} \frac{{{\rm{f}}\left( {\rm{x}} \right) - 5}}{{{\rm{x}} - 4}} = 5\]nên \[{\rm{f}}\left( 4 \right) - 5 = 0 \Rightarrow {\rm{f}}\left( 4 \right) = 5\]

Ta có:

\[\mathop {\lim }\limits_{{\rm{x}} \to 4} \frac{{{\rm{f}}\left( {\rm{x}} \right) - 5}}{{\left( {\sqrt {\rm{x}} - 2)(\sqrt {6{\rm{f}}\left( {\rm{x}} \right) + 6} + 4} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 4} \frac{{{\rm{f}}\left( {\rm{x}} \right) - 5}}{{{\rm{x}} - 4}}.\mathop {\lim }\limits_{{\rm{x}} \to 4} \frac{{\sqrt {\rm{x}} + 2}}{{\sqrt {6{\rm{f}}\left( {\rm{x}} \right) + 6} + 4}} = 5.\frac{{\sqrt 2 + 2}}{{\sqrt {6.{\rm{f}}\left( 4 \right) + 6} + 4}} = 2\]Chọn đáp án C

Đáp án cần chọn là: C

Ta có:

\[\sin {\rm{x}} - \cos {\rm{x}} = \sqrt 2 \sin \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)\]

\[\tan \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right) = \frac{{\sin \left( {{\rm{x}} - \frac{{\rm{\pi }}}{{\rm{4}}}} \right)}}{{\cos \left( {{\rm{x}} - \frac{{\rm{\pi }}}{{\rm{4}}}} \right)}}\]

\[\mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{4}} \frac{{\sin {\rm{x}} - \cos {\rm{x}}}}{{\tan \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{4}} \frac{{\sqrt 2 \sin \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right).\cos \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)}}{{\sin \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{4}} \sqrt 2 \cos \left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right) = \sqrt 2 \]

Chọn đáp án B

Đáp án cần chọn là: B

Vì\[\mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {{\rm{x}} - 2} \right) = 0 \Rightarrow \]để\[\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\sqrt {3{\rm{x}} + 3} - {\rm{m}}}}{{{\rm{x}} - 2}} = \frac{{\rm{a}}}{{\rm{b}}}\]thì\[\mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {\sqrt {3{\rm{x}} + 3} - {\rm{m}}} \right) = 0\]. Do đó x = 2 là nghiệm của phương trình \[\sqrt {3{\rm{x}} + 3} - {\rm{m}} = 0 \Rightarrow {\rm{m}} = 3\]

Với m = 3 ta được:

\[\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\sqrt {3{\rm{x}} + 3} - 3}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\left( {\sqrt {3{\rm{x}} + 3} - 3} \right)\left( {\sqrt {3{\rm{x}} + 3} + 3} \right)}}{{\left( {{\rm{x}} - 2} \right)\left( {\sqrt {3{\rm{x}} + 3} + 3} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{3{\rm{x}} - 6}}{{\left( {{\rm{x}} - 2} \right)\left( {\sqrt {3{\rm{x}} + 3} + 3} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{3\left( {{\rm{x}} - 2} \right)}}{{\left( {{\rm{x}} - 2} \right)\left( {\sqrt {3{\rm{x}} + 3} + 3} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{3}{{\left( {\sqrt {3{\rm{x}} + 3} + 3} \right)}} = \frac{1}{2}\]

\[ \Rightarrow {\rm{a}} = 1,{\rm{b}} = 2 \Rightarrow {\rm{a}} - {\rm{b}} = - 1\]Chọn đáp án C

Đáp án cần chọn là: C

Ta có \[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx}} - {\rm{5}}}}{{{\rm{x}} - 1}} = 20\]

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a + b}} - {\rm{5 = 0}}}\\{{\rm{2a + b = 20}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a = 15}}}\\{{\rm{b = }} - {\rm{10}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a + b}} - {\rm{5 = 0}}}\\{{\rm{2a + b = 20}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a = 15}}}\\{{\rm{b = }} - {\rm{10}}}\end{array}} \right.\)

\[ \Rightarrow {\rm{P = }}{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{ = 32}}5 \Rightarrow {\rm{P = }}{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{ = 3}}25\]

Đáp án cần chọn là: D

Ta có f(0) = 5n

\[\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {{\rm{mx}} + 1} - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\left( {\sqrt {{\rm{mx}} + 1} - 1} \right)\left( {\sqrt {{\rm{mx + 1}}} + 1} \right)}}{{{\rm{x}}\left( {\sqrt {{\rm{mx}} + 1} + 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\rm{m}}}{{\sqrt {{\rm{mx + 1}}} {\rm{ + 1}}}} = \frac{{\rm{m}}}{2}\]Vì hàm số liên tục tại x₀ = 0  nên \[\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right){\rm{ = f}}\left( {\rm{0}} \right) \Leftrightarrow {\rm{5n = }}\frac{{\rm{m}}}{{\rm{2}}} \Leftrightarrow {\rm{m = 10n}}\]

Chọn đáp án C

Đáp án cần chọn là: C