22 câu trắc nghiệm Toán 11 Chân trời sáng tạo Bài 2. Giới hạn của hàm số (Đúng sai - Trả lời ngắn) có đáp án

D

$\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) + 3} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) + 3 = 3 + 3 = 6$.

$\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 2x} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} 2x = 3 - 4 = - 1$.

$\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - {x^2}} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} {x^2} = 3 - 4 = - 1$.

$\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 1} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - 1 = 3 - 1 = 2$.

C

Ta có $\mathop {\lim }\limits_{x \to {x_0}} \left[ {3f\left( x \right) - 4g\left( x \right)} \right]$$= \mathop {\lim }\limits_{x \to {x_0}} 3f\left( x \right) - \mathop {\lim }\limits_{x \to {x_0}} 4g\left( x \right)$$= 3\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) - 4\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right)$$= - 6$.

D

Ta có: $\mathop {\lim }\limits_{x \to 1} \left( {2{x^2} - 3x + 1} \right) = 0$.

B

$\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{{x^3} - {x^2} - x - 2}}$$= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + x + 1} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 3}}{{{x^2} + x + 1}} = - \frac{1}{7}$.

B

$\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 2} - 2}}{{x - 2}}$$= \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{2}{{{x^2}}}} - 2}}{{x - 2}}$$= \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{2}{{{x^2}}}} - \frac{2}{x}}}{{1 - \frac{2}{x}}} = 1$.

C

Ta có $\mathop {\lim }\limits_{x \to - \infty } \frac{{1 - x}}{{3x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x} - 1}}{{3 + \frac{2}{x}}} = - \frac{1}{3}$.

A

Ta có $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 3x + 5} }}{{4x - 1}}$$= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{3}{x} + \frac{5}{{{x^2}}}} }}{{4 - \frac{1}{x}}} = - \frac{1}{4}$.

B

Ta có $\mathop {\lim }\limits_{x \to {1^ + }} \left( { - 2x + 1} \right) = - 1 < 0$, $\mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 1} \right) = 0$, $x - 1 > 0$ khi $x \to {1^ + }$.

Suy ra $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{ - 2x + 1}}{{x - 1}} = - \infty$.

C

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{x + 2}}{{x - 1}} = - \infty$ vì $\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {x + 2} \right) = 3 > 0\\\mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 0\\x - 1 < 0,\forall x < 1\end{array} \right.$.

D

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

A

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

B

Ta có $\mathop {\lim }\limits_{x \to - \infty } \left( { - 4{x^5} - 3{x^3} + x + 1} \right)$$= \mathop {\lim }\limits_{x \to - \infty } {x^5}\left( { - 4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right)$$= + \infty$.

Vì $\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - 4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) = - 4 < 0\\\mathop {\lim }\limits_{x \to - \infty } {x^5} = - \infty \end{array} \right.$.

a) $\mathop {\lim }\limits_{x \to 2} \left[ {5f\left( x \right)} \right] = 5\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5.5 = 25$.

b) $\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right).g\left( x \right)} \right] = + \infty$.

c) $\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 0$.

d) $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) - 1} - 2}}{{f\left( x \right) - 5}} = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{\left( {\sqrt {f\left( x \right) - 1} + 2} \right)\left( {f\left( x \right) - 5} \right)}}$$= \mathop {\lim }\limits_{x \to 2} \frac{1}{{\sqrt {f\left( x \right) - 1} + 2}} = \frac{1}{4}$.

Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.

a) $\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \sqrt {{x^2} + 1} = \sqrt 5$.

b) $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {{x^2} - 3x + 1} \right) = 1$.

c) $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \sqrt {{x^2} + 1} = 1$.

d) Do $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$ nên $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1$.

Đáp án: a) Sai; b) Sai; c) Đúng; d) Đúng.

a) $f\left( 8 \right) = \frac{{\sqrt {8 + 1} - 2}}{{8 - 3}} = \frac{1}{5}$.

b) $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - 2}}{{x - 3}} = \frac{1}{3}$.

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

d) $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} - 2}}{{x - 3}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} - \frac{2}{x}}}{{1 - \frac{3}{x}}} = 0$.

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

Do đó 3a + 4b = 2.

Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.

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

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

c) $\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^3} - {x^2} + x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} + 1}} = - \frac{1}{2} < 0$.

d) Để $\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = 2$ thì ax + b có nghiệm bằng 1 Û a + b = 0 Û b = −a.

Khi đó $\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{a\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{a} = - \frac{1}{a} = 2$ $\Leftrightarrow a = - \frac{1}{2} \Rightarrow b = \frac{1}{2}$.

Suy ra $a + 3b = - \frac{1}{2} + 3.\frac{1}{2} = 1$.

Đáp án: a) Đúng; b) Sai; c) Sai; d) Đúng.

a) Ta có f(2) = 2a + 2024.

b) $\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 4}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( {x + 2} \right) = 4$.

c) $\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {ax + 2024} \right) = 2a + 2024$.

d) Hàm số f(x) có giới hạn khi x → 2 khi và chỉ khi $\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)$

Û 4 = 2a + 2024 Û a = −1010.

Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.

Sau t phút bơm nước vào hồ thì lượng nước là 600 + 15t (lít) và lượng muối có được là 30.15t gam.

Nồng độ muối của nước là $C\left( t \right) = \frac{{30.15t}}{{600 + 15t}} = \frac{{30t}}{{40 + t}}$ (gam/lít).

Khi t dần về dương vô cùng, ta có $\mathop {\lim }\limits_{t \to + \infty } C\left( t \right) = \mathop {\lim }\limits_{t \to + \infty } \frac{{30t}}{{40 + t}} = \mathop {\lim }\limits_{t \to + \infty } \frac{{30}}{{\frac{{40}}{t} + 1}} = 30$ (gam/lít).

Trả lời: 30.

Ta có $\mathop {\lim }\limits_{x \to - \infty } \left[ {\left( {{m^2} - 4m + 3} \right){x^4} - x + 2025} \right] = - \infty$

$\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } {x^4}\left[ {\left( {{m^2} - 4m + 3} \right) - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right] = - \infty$.

Vì $\mathop {\lim }\limits_{x \to - \infty } {x^4} = + \infty$ và $\mathop {\lim }\limits_{x \to - \infty } \left( {{m^2} - 4m + 3 - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right) = {m^2} - 4m + 3$.

Để $\mathop {\lim }\limits_{x \to - \infty } {x^4}\left[ {\left( {{m^2} - 4m + 3} \right) - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right] = - \infty$thì m2 – 4m + 3 < 0 Û 1 < m < 3.

Mà m Î ℤ nên m = 2.

Vậy có 1 giá trị nguyên.

Trả lời: 1.

Ta có $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + ax + 2} \right) = a + 3$ và $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2{x^2} - x + 3a} \right) = 3a + 1$.

Hàm số có giới hạn khi x → 1 khi và chỉ khi $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$ Û a + 3 = 3a + 1 Û a = 1.

Trả lời: 1.

$\mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x - 3}}{{\sqrt {5x + 1} - 4}}$$= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {x - 3} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{{5\left( {x - 3} \right)}}$

$= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{5} = \frac{{56}}{5} = 11{,}2$.

Trả lời: 11,2.

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

Suy ra $a = 1;\,b = 2$.

$\sqrt a + b + 2018 = 1 + 2 + 2018 = 2021$.

Trả lời: 2021.