20 câu trắc nghiệm Toán 11 Kết nối tri thức Bài 16. Giới hạn của hàm số có đáp án

D

Ta có $\mathop {\lim }\limits_{{\rm{x}} \to \pm \infty } {\rm{c}} = {\rm{c}}$.

A

$\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{ + g}}\left( {\rm{x}} \right)} \right] = {\rm{L + M}}$.

C

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

C

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

A

Khi ${\rm{x}} \to {3^ + }$thì x > 3

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

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

B

Ta có 3 – x > 0 với mọi x < 3

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

D

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

D

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

A

Ta có:

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

$\mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \left( {5 - {\rm{x}}} \right) = 0$ và 5 – x > 0 khi x → 5$^{−}$.

$\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {5^ - }} \frac{{12 - {{\rm{x}}^2}}}{{5 - {\rm{x}}}} = - \infty$.

B

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

Vì $\mathop {\lim }\limits_{x \to {3^ + }} \left( {x + 2} \right) = 3 + 2 = 5 > 0$ và $\mathop {\lim }\limits_{x \to {3^ + }} \left( {x - 3} \right) = 0$; x – 3 > 0 khi x → 3$^{+}$.

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

$\Leftrightarrow$ 4 = 2a + 2024 $\Leftrightarrow$ a = −1010.

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) Với m = −1 thì $\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1 - 2} \right) = 4 - 1 - 2 = 1$.

b) $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \sqrt {x + 7} = \sqrt {3 + 7} = \sqrt {10}$.

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

$\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7} = 3$.

Để tồn tại $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ thì $\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)$ Û 3 + 2m = 3 Û m = 0.

d) $\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7} = 3$.

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

a) $\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \left( {x - 2} \right) = - 4$ .

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

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

d) Vì $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right)$ nên không tồn tại giới hạn của hàm số khi x → −1.

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

a) $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1$.

b) $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right]$.

c) Khi b = 2 thì $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + 2x} \right)$$= \mathop {\lim }\limits_{x \to - \infty } \frac{{ax + 1}}{{\sqrt {4{x^2} + ax + 1} - 2x}}$$= \mathop {\lim }\limits_{x \to - \infty } \frac{{a + \frac{1}{x}}}{{ - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} - 2}}$$= - \frac{a}{4}$.

d) $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right]$.

Nếu b ≠ 2: $\mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right] = \left\{ \begin{array}{l} - \infty \;khi\;b > 2\\ + \infty \;khi\;b < 2\end{array} \right.$ mâu thuẫn với giải thiết.

Vậy b = 2.

Khi đó $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = - \frac{a}{4}$.

Mà $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = - 1$ nên $- \frac{a}{4} = - 1 \Rightarrow a = 4$.

Vậy a = 4; b = 2. Do đó P = a$^{2}$ – 2b$^{3}$ = 0.

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

$\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 7ax + 5} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{7ax + 5}}{{\sqrt {{x^2} + 7ax + 5} - x}}$$= \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {7a + \frac{5}{x}} \right)}}{{x\left( { - \sqrt {1 + \frac{{7a}}{x} + \frac{5}{{{x^2}}}} - 1} \right)}} = - \frac{{7a}}{2}$.

Mà $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 7ax + 5} + x} \right) = - 3$ nên $- \frac{{7a}}{2} = - 3$$\Leftrightarrow a = \frac{6}{7} \approx 0{,}86$.

Trả lời: 0,86.

$\mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{4{x^2} - 3x + 2}}{{x + 2}} - 2ax + b} \right) = 0$$\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \left[ {\left( {4 - 2a} \right)x + b - 11 + \frac{{24}}{{x + 2}}} \right] = 0$

$\Leftrightarrow \left\{ \begin{array}{l}4 - 2a = 0\\ - 11 + b = 0\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 11\end{array} \right.$.

Vậy 2a – 3b = −29.

Trả lời: −29.

$\mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{3 - \sqrt {9 + x} }}$$= \mathop {\lim }\limits_{x \to 0} \frac{{4x\left( {3 + \sqrt {9 + x} } \right)}}{{9 - \left( {9 + x} \right)}}$$= \mathop {\lim }\limits_{x \to 0} \left[ { - 4\left( {3 + \sqrt {9 + x} } \right)} \right] = - 24$.

Trả lời: −24.

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 m$^{2}$ – 4m + 3 < 0 Û 1 < m < 3.

Mà m Î ℤ nên m = 2.

Vậy có 1 giá trị nguyên.

Trả lời: 1.