Trắc nghiệm Giới hạn của hàm số lớp 11 (có đáp án)

Lời giải:

Ta có $\mathop {\lim }\limits_{{\rm{x}} \to \pm \infty } {\rm{c}} = {\rm{c}}$⇒ Chọn đáp án D.

Đáp án cần chọn là: D

Lời giải:

Chọn đáp án A

Đáp án cần chọn là: A

Ta có $\mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {2 + {\rm{x}}} \right) = 2 + 2 = 4$

Chọn đáp án C

Đáp án cần chọn là: C

$${\rm{2g(x)] = }}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{f}}\left( {\rm{x}} \right) + \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{2g}}\left( {\rm{x}} \right) = 1 + 2.4 = 9$$

Đáp án cần chọn là: C

Lời giải:

Chọn đáp án A

Đáp án cần chọn là: A

Chọn đáp án A

Đáp án cần chọn là: A

Chọn đáp án B

Đáp án cần chọn là: B

$$\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{{{\rm{x}}^3} + 3{{\rm{x}}^2} + 4}}{{2{{\rm{x}}^3}}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{{{\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{3}}}}}}^3} + {{\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{3}}}}}}^2} + \frac{{\rm{4}}}{{{{\rm{x}}^{\rm{3}}}}}}}{{\frac{{{\rm{2}}{{\rm{x}}^{\rm{3}}}}}{{{{\rm{x}}^{\rm{3}}}}}}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{1 + \frac{1}{{\rm{x}}} + \frac{4}{{{{\rm{x}}^3}}}}}{2} = \frac{1}{2}$$

Chọn đáp án D

Đáp án cần chọn là: D

Ta có $\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {\frac{{\left( {{{\rm{x}}^2} + 3} \right){\rm{x}}}}{{{{\rm{x}}^3} - 1}}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {\frac{{\left( {{2^2} + 3} \right) \times 2}}{{{2^3} - 1}}} = \sqrt {\frac{{14}}{7}} = \sqrt 2 = \frac{2}{{\sqrt 2 }}$

Chọn đáp án A

Đáp án cần chọn là: A

Ta có

$$\mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} \left( {{{\rm{x}}^2} - 3} \right) = {2^2} - 3 = 1$$

$$\mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} \left( {{\rm{x}} - 1} \right) = 2 - 1 = 1$$

$$\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{f}}\left( {\rm{x}} \right) = 1$$

Chọn đáp án B

Đáp án cần chọn là: B

Khi ${\rm{x}} \to {3^ + }$thì x > 3

Nên ${\rm{x}} - 3 > 0 \Rightarrow \left| {{\rm{x}} - 3} \right| = {\rm{x}} - 3$

$$\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{\left| {{\rm{x}} - 3} \right|}}{{3{\rm{x}} - 9}} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{{\rm{x}} - 3}}{{3{\rm{x}} - 9}} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{{\rm{x}} - 3}}{{3.\left( {{\rm{x}} - 3} \right)}} = \frac{1}{3}$$

Chọn đáp án A

Đáp án cần chọn là: A

$$\mathop {\lim }\limits_{{\rm{x}} \to - 3} \left| {\frac{{ - {{\rm{x}}^2} - {\rm{x}} + 6}}{{{{\rm{x}}^2} + 3{\rm{x}}}}} \right| = \mathop {\lim }\limits_{{\rm{x}} \to - 3} \left| {\frac{{ - ({\rm{x}} + 3)({\rm{x}} - 2)}}{{{\rm{x}}({\rm{x}} + 3)}}} \right| = \mathop {\lim }\limits_{{\rm{x}} \to - 3} \left| {\frac{{ - ({\rm{x}} - 2)}}{{\rm{x}}}} \right| = \left| {\frac{{ - ( - 3 - 2)}}{{ - 3}}} \right| = \frac{5}{3}$$

Chọn đáp án B

Đáp án cần chọn là: B

Ta có 3 – x > 0 với mọi x < 3

$$\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{3 - {\rm{x}}}}{{\sqrt {27 - {{\rm{x}}^3}} }} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{3 - {\rm{x}}}}{{\sqrt {\left( {3 - {\rm{x}}} \right)\left( {9 + 3{\rm{x}} + {{\rm{x}}^2}} \right)} }} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{\sqrt {3 - {\rm{x}}} }}{{\sqrt {9 + 3{\rm{x}} + {{\rm{x}}^2}} }} = \frac{{\sqrt {3 - 3} }}{{\sqrt {9 + 3.3 + {3^2}} }} = 0$$

Chọn đáp án B

Đáp án cần chọn là: B

$$\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {{{\left| {\rm{x}} \right|}^5} + {\rm{x}} + 1} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left[ {{{\left( { - {\rm{x}}} \right)}^5} + {\rm{x}} + 1} \right] = + \infty$$

Chọn đáp án A

Đáp án cần chọn là: A

$$\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \frac{{12 - {{\rm{x}}^2}}}{{5 - {\rm{x}}}} = \frac{{\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {12 - {{\rm{x}}^2}} \right)}}{{\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {5 - {\rm{x}}} \right)}}$$

Ta có:

$$\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {12 - {{\rm{x}}^2}} \right) = - 13 < 0$$

$$\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {5 - {\rm{x}}} \right) > 0$$

$$\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \frac{{12 - {{\rm{x}}^2}}}{{5 - {\rm{x}}}} = \frac{{\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {12 - {{\rm{x}}^2}} \right)}}{{\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {5 - {\rm{x}}} \right)}} = - \infty$$

Chọn đáp án A

Đáp án cần chọn là: A

Ta có

$${\rm{A = }}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{{{\rm{x}}^{\rm{n}}} - {\rm{1}}}}{{{{\rm{x}}^{\rm{m}}} - {\rm{1}}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{{\rm{(x}} - {\rm{1)(}}{{\rm{x}}^{{\rm{n}} - {\rm{1}}}}{\rm{ + }}{{\rm{x}}^{{\rm{n}} - {\rm{2}}}}{\rm{ + }}{{\rm{x}}^{{\rm{n}} - {\rm{3}}}}{\rm{ + }}...{\rm{ + x + 1)}}}}{{{\rm{(x}} - {\rm{1)(}}{{\rm{x}}^{{\rm{m}} - {\rm{1}}}}{\rm{ + }}{{\rm{x}}^{{\rm{m}} - {\rm{2}}}}{\rm{ + }}{{\rm{x}}^{{\rm{m}} - {\rm{3}}}}{\rm{ + }}...{\rm{ + x + 1)}}}}$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{{\rm{x}}^{{\rm{n}} - {\rm{1}}}}{\rm{ + }}{{\rm{x}}^{{\rm{n}} - {\rm{2}}}}{\rm{ + }}{{\rm{x}}^{{\rm{n}} - {\rm{3}}}}{\rm{ + }}...{\rm{ + x + 1}}}}{{{{\rm{x}}^{{\rm{m}} - {\rm{1}}}}{\rm{ + }}{{\rm{x}}^{{\rm{m}} - {\rm{2}}}}{\rm{ + }}{{\rm{x}}^{{\rm{m}} - 3}}{\rm{ + }}...{\rm{ + x + 1}}}} = \frac{{1 + 1 + 1 + ... + 1 + 1}}{{1 + 1 + 1 + ... + 1 + 1}} = \frac{{\rm{n}}}{{\rm{m}}}$$

Chọn đáp án D

Đáp án cần chọn là: D

$$\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {{\rm{x}} + 5} - \sqrt {{\rm{x}} - 6} } \right) = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \frac{{{\rm{x}} + 5 - {\rm{x}} + 6}}{{\sqrt {{\rm{x}} + 5} + \sqrt {{\rm{x}} - 6} }} = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \frac{{11}}{{\sqrt {{\rm{x}} + 5} + \sqrt {{\rm{x}} - 6} }} = \frac{{11}}{\infty } = 0$$

Chọn đáp án D

Đáp án cần chọn là: D

$$\mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi }}}{{\rm{2}}}}^ + }} \left[ {\left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{x}}} \right).\tan {\rm{x}}} \right] = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi }}}{{\rm{2}}}}^ + }} \left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{x}}} \right).\frac{{\sin {\rm{x}}}}{{\cos {\rm{x}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{\pi }{2}}^ + }} \left[ {\left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{x}}} \right).\frac{{\sin {\rm{x}}}}{{\sin \left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{x}}} \right)}}} \right] = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi }}}{{\rm{2}}}}^ + }} \left( {\sin {\rm{x}}} \right) = 1$$

Chọn đáp án B

Đáp án cần chọn là: B

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 x$_{0}$= 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

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

$\Rightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a + b}} - {\rm{5 = 0}}}\\{{\rm{2a + b = 20}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a = 15}}}\\{{\rm{b = }} - {\rm{10}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a + b}} - {\rm{5 = 0}}}\\{{\rm{2a + b = 20}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a = 15}}}\\{{\rm{b = }} - {\rm{10}}}\end{array}} \right.$

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

Đáp án cần chọn là: D

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:

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

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

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

Chọn đáp án B

Đáp án cần chọn là: B

Vì$\mathop {\lim }\limits_{{\rm{x}} \to 4} \frac{{{\rm{f}}\left( {\rm{x}} \right) - 5}}{{{\rm{x}} - 4}} = 5$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$\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {2020{{\rm{x}}^2} + {\rm{x}} + 3} - \sqrt {2021{{\rm{x}}^2} + 2} } \right) = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left[ {{\rm{x}}\left( {\sqrt {2020 + \frac{1}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right)} \right]$

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

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

Đáp án cần chọn là: A

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

Ta có $\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2 + 2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}}$

$$= \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} + \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}}$$

Tính${\rm{I}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{{\rm{x}}^2} + {\rm{x}} + 2 - 4}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}}$

$$= \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 2} \right)}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{x}} + 2}}{{\sqrt 2 \left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \frac{3}{{4\sqrt 2 }}$$

và${\rm{J}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{8 - 7{\rm{x}} - 1}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}}$

$$= \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}} = \frac{{ - 7}}{{12\sqrt 2 }}$$

Do đó$\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}} = \frac{{\sqrt 2 }}{{12}}$

Suy ra a = 1, b = 12, c = 0. Vậy a + b + c = 13.

Đáp án cần chọn là: A

Đường thẳng ${\rm{\Delta }}:{\rm{y = ax + 6b}}$ đi qua điểm M(3; 42) nên:

$${\rm{3a + 6b}} = 42 \Rightarrow {\rm{a}} + 2{\rm{b}} = 14$$

Ta có:

$$\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {{\rm{36}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5ax + 1}}} - {\rm{6x + b}}} \right)$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\frac{{\left( {\sqrt {{\rm{36}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5ax + 1}}} - 6{\rm{x}}} \right)\left( {\sqrt {{\rm{36}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5ax + 1}}} {\rm{ + 6x}}} \right)}}{{\sqrt {{\rm{36}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5ax + 1}}} {\rm{ + 6x}}}} + {\rm{b}}} \right)$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\frac{{{\rm{5ax + 1}}}}{{\sqrt {{\rm{36}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5ax + 1}}} {\rm{ + 6x}}}} + {\rm{b}}} \right)$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\frac{{{\rm{5a + }}\frac{{\rm{1}}}{{\rm{x}}}}}{{\sqrt {{\rm{36 + }}\frac{{{\rm{5a}}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}}}{\rm{ + 6}}} }}{\rm{ + b}}} \right) = \frac{{{\rm{5a}}}}{{{\rm{12}}}}{\rm{ + b}}$$

Do đó $\frac{{{\rm{5a}}}}{{{\rm{12}}}}{\rm{ + b = }}\frac{{{\rm{20}}}}{{\rm{3}}} \Rightarrow {\rm{5a + 12b = }}80$. Ta có hệ: $\left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{5a + 12b = 80}}}\\{{\rm{a + 2b = 14}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a = 4}}}\\{{\rm{b = 5}}}\end{array}} \right.$.

Vậy ${\rm{T = }}{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{ = 41}}$

Đáp án cần chọn là: C

$$\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + {\rm{3x}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}} - 3{\rm{x}}} \right)$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) + \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)$$

Ta có :

$$\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{ - {\rm{ax}}}}{{\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} - 3{\rm{x}}}} = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\rm{a}}}{{\sqrt {9 - \frac{{\rm{a}}}{{\rm{x}}}} + 3}} = \frac{{\rm{a}}}{{\rm{6}}}$$

Ta có :

$$\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - {\rm{3x}}} \right)$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)\left( {\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}} \right)}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}$$

$$= \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}}{{\sqrt[3]{{{{\left( {{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{3}}}}}} \right)}^2}}} + 3.\sqrt[3]{{{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}} + 9}} = \frac{{\rm{b}}}{{{\rm{27}}}}$$

Do đó$\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}}{\rm{ = }}\frac{{\rm{7}}}{{{\rm{27}}}}$

Áp dụng bất đẳng thức Cauchy cho 2 số dương, ta có :$\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}} \ge 2\sqrt {\frac{{\rm{a}}}{{\rm{6}}}{\rm{.}}\frac{{\rm{b}}}{{{\rm{27}}}}}$

$$\Rightarrow \frac{7}{{27}} \ge \frac{2}{{9\sqrt 2 }}\sqrt {{\rm{a}}{\rm{.b}}} \Rightarrow {\rm{ab}} \le \frac{{49}}{{18}}$$

Đẳng thức xảy ra khi$\left\{ {\begin{array}{cccccccccccccccccccc}{\frac{{\rm{a}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{b}}}{{{\rm{2}}7}}}\\{\frac{{\rm{a}}}{{\rm{6}}} + \frac{{\rm{b}}}{{{\rm{2}}7}} = \frac{{\rm{7}}}{{{\rm{2}}7}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{a}} = \frac{7}{9}}\\{{\rm{b}} = \frac{7}{2}}\end{array}} \right.$

Vậy giá trị lớn nhất của ab bằng $\frac{{49}}{{18}}$.

Đáp án cần chọn là: B