22 câu Trắc nghiệm Toán 11 Cánh diều Bài 2. Giới hạn của hàm số (Đúng-sai, trả lời ngắn) có đáp án

D

$\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right]$$= \mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = - 3 + 5 = 2$.

D

$\mathop {\lim }\limits_{x \to 3} \left[ {4x - 3f\left( x \right)} \right]$$= \mathop {\lim }\limits_{x \to 3} \left( {4x} \right) - 3\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 12 - 3.2 = 6$.

A

$\mathop {\lim }\limits_{x \to 2} \frac{{5x - 5}}{{2x}}$$= \frac{{5.2 - 5}}{{2.2}} = \frac{5}{4}$.

A

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

D

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

A

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

B

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {10 - {x^2}} - 3}}{{1 - {x^2}}}$$= \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2}}}{{\left( {1 - {x^2}} \right)\left( {\sqrt {10 - {x^2}} + 3} \right)}}$$= \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {10 - {x^2}} + 3}} = \frac{1}{6}$.

C

Vì $\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$$\ne \mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right)$ nên không tồn tại $\mathop {\lim }\limits_{x \to 5} f\left( x \right)$.

C

Ta có $\mathop {\lim }\limits_{x \to {1^ - }} \left( {x - 1} \right) = 0$ và x – 1 < 0 khi x → 1−.

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

C

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

A

Ta có $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {b{x^2} - 2x} \right) = b - 2$; $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {ax + 5} \right) = a + 5$.

Vì $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 3$ nên $\left\{ \begin{array}{l}b - 2 = 3\\a + 5 = 3\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}b = 5\\a = - 2\end{array} \right.$. Do đó a + b = 3.

C

Đặt $g\left( x \right) = \frac{{f\left( x \right) - 2}}{{x + 1}}$$\Rightarrow f\left( x \right) = \left( {x + 1} \right)g\left( x \right) + 2$.

Ta cần tính $\mathop {\lim }\limits_{x \to - 1} f\left( x \right)$.

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

Ta có $\mathop {\lim }\limits_{x \to - 1} \frac{{{f^2}\left( x \right) + f\left( x \right) - 6}}{{x + 1}}$$= \mathop {\lim }\limits_{x \to - 1} \frac{{\left[ {f\left( x \right) + 3} \right]\left[ {f\left( x \right) - 2} \right]}}{{x + 1}}$

$= \mathop {\lim }\limits_{x \to - 1} \frac{{\left[ {f\left( x \right) - 2} \right]}}{{x + 1}}.\mathop {\lim }\limits_{x \to - 1} \left[ {f\left( x \right) + 3} \right] = 2024.\left( {2 + 3} \right) = 2024.5 = 10120$.

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.

a) $\mathop {\lim }\limits_{x \to 2} 3f\left( x \right) = 3\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3.2024 = 6072$.

b) $\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{4} = \frac{{2024}}{4} = 506$.

c) $\mathop {\lim }\limits_{x \to 2} \sqrt {f\left( x \right)} = \sqrt {2024} = 2\sqrt {506}$.

d) $\mathop {\lim }\limits_{x \to 2} \left[ {100x - \frac{1}{2}f\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 2} 100x - \frac{1}{2}\mathop {\lim }\limits_{x \to 2} f\left( x \right)$$= 200 - \frac{1}{2}.2024 = - 812$.

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

a) Với m = 1 thì $\mathop {\lim }\limits_{x \to - 1} g\left( x \right) = \mathop {\lim }\limits_{x \to - 1} \frac{{3x + 1}}{{ - x + 1}} = - 1$.

b) Với m = 1 thì $\mathop {\lim }\limits_{x \to {1^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{3x + 1}}{{ - x + 1}} = + \infty$

vì $\mathop {\lim }\limits_{x \to {1^ - }} \left( {3x + 1} \right) = 4;\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x + 1} \right) = 0$ và $- x + 1 > 0$khi x → 1−.

c) Với m = 1 thì $\mathop {\lim }\limits_{x \to {1^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{3x + 1}}{{ - x + 1}} = - \infty$

Vì $\mathop {\lim }\limits_{x \to {1^ + }} \left( {3x + 1} \right) = 4;\mathop {\lim }\limits_{x \to {1^ + }} \left( { - x + 1} \right) = 0$ và $- x + 1 < 0$khi x → 1+.

d) $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {5x + 1} - 4}}{{27 - {x^3}}}$$= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {5x + 1 - 16} \right)}}{{\left( {27 - {x^3}} \right)\left( {\sqrt {5x + 1} + 4} \right)}}$

$= \mathop {\lim }\limits_{x \to 3} \frac{{5\left( {x - 3} \right)}}{{\left( {3 - x} \right)\left( {9 + 3x + {x^2}} \right)\left( {\sqrt {5x + 1} + 4} \right)}}$$= \mathop {\lim }\limits_{x \to 3} \frac{{ - 5}}{{\left( {9 + 3x + {x^2}} \right)\left( {\sqrt {5x + 1} + 4} \right)}}$$= \frac{{ - 5}}{{\left( {9 + 3.3 + {3^2}} \right)\left( {\sqrt {5.3 + 1} + 4} \right)}} = - \frac{5}{{216}}$ Þ $a = - 5$.

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

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

d) Hàm số đã cho có giới hạn tại x0 = −1 khi $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right)$$\Leftrightarrow m = - 3 - \sqrt 2$.

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

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 + \infty } \left( {\frac{{{x^2} + 3x + 1}}{{x + 1}} + ax + b} \right) = 1$$\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\left( {a + 1} \right){x^2} + \left( {a + b + 3} \right)x + b + 1}}{{x + 1}}} \right) = 1$

$\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\left( {a + 1} \right)x + \left( {a + b + 3} \right) + \frac{{b + 1}}{x}}}{{1 + \frac{1}{x}}}} \right) = 1$ $\Leftrightarrow \left\{ \begin{array}{l}a + 1 = 0\\a + b + 3 = 1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}a = - 1\\b = - 1\end{array} \right.$.

Do đó T = 2024a – 4049b = 2024.(−1) – 4049.(−1) = 2025.

Trả lời: 2025.

Chi phí trung bình để sản xuất một sản phẩm là $\overline C \left( t \right) = \frac{{200t + 1000000}}{t}$.

Ta có $\mathop {\lim }\limits_{t \to + \infty } \overline C \left( t \right) = \mathop {\lim }\limits_{t \to + \infty } \frac{{200t + 1000000}}{t} = \mathop {\lim }\limits_{t \to + \infty } \left( {200 + \frac{{1000000}}{t}} \right) = 200$.

Vậy khi số sản phẩm càng lớn thì chi phí trung bình để sản xuất một sản phẩm tiến dần đến 200 nghìn đồng.

Trả lời: 200.

$A = \mathop {\lim }\limits_{x \to 1} \frac{{x - m}}{{x + 2}} = - 2025$ $\Leftrightarrow \frac{{1 - m}}{3} = - 2025$$\Leftrightarrow m = 6076$.

Trả lời: 6076.