Đề kiểm tra Giới hạn của hàm số lớp 11 (có lời giải) - Đề 2

Chọn A

Ta có: $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 2{\rm{x}} + 3}}{{x + 1}} = \frac{{{1^2} - 2.1 + 3}}{{1 + 1}} = 1$.

Chọn A

Dễ thấy $\mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{{x - 1}} = \frac{{2 + 2}}{{2 - 1}} = 4$

Chọn B

$$\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 4} \right| = \left| {3 - 4} \right| = 1$$

Chọn D

Ta có $\mathop {\lim }\limits_{x \to 3} \left[ {f\left( x \right) + 4x - 1} \right] = 9$.

Chọn B

Vì $\sin \frac{\pi }{2} = 1$ nên $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sin x}}{x} = \frac{2}{\pi }$.

Ta có

$I = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {3x + 1} - 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{6x}}{{x\left( {\sqrt {3x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{6}{{\sqrt {3x + 1} + 1}} = 3$.

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

Khi đó $I - J = 6$.

Chọn A

Xét $\mathop {\lim }\limits_{x \to {4^ - }} \frac{{2x - 1}}{{4 - x}}$

Ta có $\mathop {\lim }\limits_{x \to {4^ - }} \left( {2x - 1} \right) = 7 > 0$, $\mathop {\lim }\limits_{x \to {4^ - }} \left( {4 - x} \right) = 0$ và $4 - x > 0$ với mọi $x < 4$

Do đó $\mathop {\lim }\limits_{x \to {4^ - }} \frac{{2x - 1}}{{4 - x}} = + \infty$.

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

Lời giải

Ta có:

$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {x + 4} - 2}}{x}$$= \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left( {x + 4} \right) - {2^2}}}{{x\left( {\sqrt {x + 4} + 2} \right)}}$$= \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{x\left( {\sqrt {x + 4} + 2} \right)}}$$= \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{\sqrt {x + 4} + 2}} = \frac{1}{4}$.

$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {mx + m + \frac{1}{4}} \right) = m + \frac{1}{4}$

Hàm số đã cho có giới hạn tại $x = 0$ khi và chỉ khi $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$

$\Rightarrow F = \left( { - \frac{5}{2}; - 2;0} \right)$.

Chọn C

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

Chọn B

$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{{{\left( {4x + 1} \right)}^3}{{\left( {2x + 1} \right)}^4}}}{{{{\left( {3 + 2x} \right)}^7}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{{\left( {4 + \frac{1}{x}} \right)}^3}{{\left( {2 + \frac{1}{x}} \right)}^4}}}{{{{\left( {\frac{3}{x} + 2} \right)}^7}}} = {2^3} = 8$.

a) Đúng

b) Sai

c) Đúng

d) Đúng

Chọn B

Ta có: $\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = + \infty$ do $\mathop {\lim }\limits_{x \to {0^ + }} x = 0$và $x > 0$. Vậy đáp án A đúng.

Suy ra đáp án B sai.

Các đáp án C và D đúng. Giải thích tương tự đáp án A.

a) Sai

b) Đúng

c) Đúng

d) Sai

a) Ta có $\mathop {\lim }\limits_{x \to 3} f(x) = \sqrt 5$

b) Xét dãy số $\left( {{x_n}} \right)$ bất kì sao cho ${x_n} < 2$ và ${x_n} \to 2$, ta có: $f\left( {{x_n}} \right) = 1 - x_n^2$.

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

c) Xét dãy số $\left( {{x_n}} \right)$ bất kì sao cho ${x_n} > 2$ và ${x_n} \to 2$, ta có $f\left( {{x_n}} \right) = \sqrt {{x_n} + 2}$.

Khi đó: $\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \lim f\left( {{x_n}} \right) = \sqrt {2 + 2} = 2$.

d) Vì $\mathop {\lim }\limits_{x \to {2^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {2^ + }} f(x)$ (hay $- 3 \ne 2$) nên không tồn tại $\mathop {\lim }\limits_{x \to 2} f(x)$.

a) Đúng

b) Sai

c) Sai

d) Sai

a) $\mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{x + 1}} = \frac{2}{{2 + 1}} = \frac{2}{3}$.

b) $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{2x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {(2x - 1) \cdot \frac{1}{{x - 1}}} \right] = + \infty$ (do $\mathop {\lim }\limits_{x \to {1^ + }} (2x - 1) = 1$ và $\mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{x - 1}} = + \infty$).

c) $\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} - 3x}}{{{x^2} - 6x + 9}} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{x(x - 3)}}{{{{(x - 3)}^2}}} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ - }} \left( {x\frac{1}{{x - 3}}} \right) = - \infty$,

do $\mathop {\lim }\limits_{x \to {3^ - }} x = 3$ và $\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}} = - \infty$.

d)

${\rm{ }}\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {{x^3} - 1} \right)\left( {\sqrt {\frac{x}{{{x^2} - 1}}} } \right)} \right] = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {(x - 1)\left( {{x^2} + x + 1} \right)\sqrt {\frac{x}{{(x - 1)(x + 1)}}} } \right]}\\{ = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {{x^2} + x + 1} \right)\sqrt {\frac{{x{{(x - 1)}^2}}}{{(x - 1)(x + 1)}}} } \right] = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {{x^2} + x + 1} \right)\sqrt {\frac{{x(x - 1)}}{{x + 1}}} } \right] = 3 \cdot \sqrt {\frac{0}{2}} = 0.}\end{array}$

a) Đúng

b) Đúng

c) Đúng

d) Sai

a) $\mathop {\lim }\limits_{x \to {2^ + }} (\sqrt {x + 2} - 1) = \sqrt {2 + 2} - 1 = 1$.

b) $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{4x - 3}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {(4x - 3) \cdot \frac{1}{{x - 1}}} \right] = + \infty$ vì $\mathop {\lim }\limits_{x \to {1^ + }} (4x - 3) = 1,\mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{x - 1}} = + \infty$.

c)

${\rm{ }}\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{{x - 2}} - \frac{1}{{{x^2} - 4}}} \right)}&{ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 2 - 1}}{{(x - 2)(x + 2)}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 1}}{{(x - 2)(x + 2)}}}\\{}&{ = \mathop {\lim }\limits_{x \to {2^ - }} \left[ {\frac{{x + 1}}{{x + 2}} \cdot \frac{1}{{(x - 2)}}} \right] = - \infty ,{\rm{ do }}\left\{ {\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 1}}{{x + 2}} = \frac{3}{4}}\\{\mathop {\lim }\limits_{x \to {2^ - }} \frac{1}{{x - 2}} = - \infty }\end{array}} \right.}\end{array}$

d) $\mathop {\lim }\limits_{x \to - {1^ - }} \frac{{|x + 1|}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{ - x - 1}}{{(x - 1)(x + 1)}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{ - 1}}{{x - 1}} = \frac{1}{2}$.

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

Để tồn tại $\mathop {\lim }\limits_{x \to 1} f(x)$ thì $\sqrt {1 + a} = 3$, suy ra $a = 8$.

Xét hàm số $f\left( x \right) = a{x^2} + bx + c$.

Hàm số $f\left( x \right) = a{x^2} + bx + c$ liên tục trên $\mathbb{R}$.

Ta có $f\left( {\frac{4}{5}} \right) = \frac{{16}}{{25}}a + \frac{4}{5}b + c$ nên $\frac{{75}}{4}f\left( {\frac{4}{5}} \right) = 12a + 15b + \frac{{75}}{4}c$.

$f\left( 0 \right) = c$ nên $\frac{5}{4}f\left( 0 \right) = \frac{5}{4}c$.

Do đó $\frac{{75}}{4}f\left( {\frac{4}{5}} \right) + \frac{5}{4}f\left( 0 \right) = 12a + 15b + 20c = 0$.

Suy ra $f\left( {\frac{4}{5}} \right)$, $f\left( 0 \right)$ trái dấu hoặc cả hai đều bằng 0.

Vậy phương trình $a{x^2} + bx + c = 0$ luôn có nghiệm thuộc $\left[ {0;\frac{4}{5}} \right]$.