Đề kiểm tra Giới hạn của hàm số lớp 11 (có lời giải) - Đề 3

Chọn C

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

Chọn A

Ta có: Với $x = - 2$; ${x^2} + x + 4 \ne 0$

Nên $A = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 1}}{{{x^2} + x + 4}} = \frac{{\left( { - 2} \right) + 1}}{{{{\left( { - 2} \right)}^2} + \left( { - 2} \right) + 4}} = - \frac{1}{6}$.

Chọn D

Ta có: $\left\{ {\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to {a^ - }} 1 = 1 > 0}\\{\mathop {\lim }\limits_{x \to {a^ - }} \left( {1 - a} \right) = 0}\\{x - a < 0{\rm{ khi }}x \to {a^ - }}\end{array}} \right.$

Vậy $\mathop {\lim }\limits_{x \to {a^ - }} \frac{1}{{x - a}} = - \infty$.

Chọn C

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

Chọn C

Ta có: $\mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} + 1} \right) = 2 > 0;\;\mathop {\lim }\limits_{x \to {1^ - }} \left( {x - 1} \right) = 0$ và $x - 1 < 0,\;\forall x < 1\;$(do $x \to {1^ - }$)

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

Chọn C

Ta có: + $\mathop {\lim }\limits_{x \to - 1} {\mkern 1mu} f(x) = 4 > 0$.

+ $\mathop {\lim }\limits_{x \to - 1} {\mkern 1mu} {\left( {x + 1} \right)^4} = 0$ và với $\forall x \ne - 1$ thì ${\left( {x + 1} \right)^4} > 0$.

Suy ra $\mathop {\lim }\limits_{x \to - 1} {\mkern 1mu} \frac{{f(x)}}{{{{\left( {x + 1} \right)}^4}}} = + \infty$.

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

Chọn D

$\mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{4{x^2} - 3x + 1}}{{x + 2}} - ax - b} \right) = 0$$\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\left( {4 - a} \right)x - b - 11 + \frac{{23}}{{x + 2}}} \right) = 0$$\Leftrightarrow \left\{ \begin{array}{l}4 - a = 0\\ - 11 - b = 0\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}a = 4\\b = - 11\end{array} \right.$

Chọn D

Ta có$\mathop {\lim }\limits_{x \to + \infty } \frac{{ax + \sqrt {{x^2} - 3x + 5} }}{{2x - 7}} = 2$$\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{a + \sqrt {1 - \frac{3}{x} + \frac{5}{{{x^2}}}} }}{{2 - \frac{7}{x}}} = 2$$\Leftrightarrow \frac{{a + 1}}{2} = 2$$\Leftrightarrow \frac{{a + 1}}{2} = 3$.

$\Leftrightarrow a + 1 = 6 \Leftrightarrow a = 5$

Lời giải

Chọn B

Xét mọi dãy số $\left( {{x_n}} \right)$ sao cho $\lim {x_n} = + \infty \Rightarrow \lim \frac{1}{{{x_n}}} = 0$

Ta có $\mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\sin x}}{x}} \right) = \lim \left( {\frac{{\sin {x_n}}}{{{x_n}}}} \right)$

Ta có $\left| {\frac{{\sin {x_n}}}{{{x_n}}}} \right| \le \frac{1}{{{x_n}}}$ mà $\mathop {\lim }\limits_{} \left( {\frac{1}{{{x_n}}}} \right) = 0$ nên $\left| {\frac{{\sin {x_n}}}{{{x_n}}}} \right|$ nhỏ hơn một số dương bé tùy ý kể từ số hạng nào đó trở đi

Theo định nghĩa dãy số có giới hạn $0$ ta có $\lim \left( {\frac{{\sin {x_n}}}{{{x_n}}}} \right) = 0$

Vậy $\mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\sin x}}{x}} \right) = 0$

a) Đúng

b) Sai

c) Đúng

d) Sai

a) $\mathop {\lim }\limits_{x \to + \infty } \left( {{x^2} + 3} \right) = \mathop {\lim }\limits_{x \to + \infty } {x^2}\left( {1 + \frac{3}{{{x^2}}}} \right) = + \infty$, do $\mathop {\lim }\limits_{x \to + \infty } {x^2} = + \infty$ và $\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{3}{{{x^2}}}} \right) = 1$.

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

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

c) $\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x \cdot \frac{1}{x}}}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x}}}{{1 + \frac{2}{x}}} = 0$.

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

a) Sai

b) Sai

c) Đúng

d) Đúng

a) $\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} - 2x} \right) = {3.2^2} - 2.2 = 8$

b) $\mathop {\lim }\limits_{x \to - 2} \frac{{4{x^2} + 2x + 1}}{{x - 4}} = - \frac{{13}}{6}$

c) $\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - {x^2} + 2x - 24}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{(x - 3)\left( {{x^2} + 2x + 8} \right)}}{{(x - 3)(x + 3)}} = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} + 2x + 8}}{{x + 3}} = \frac{{23}}{6}$

d) $\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} - x - 14}}{{{x^2} - 7x - 18}} = \mathop {\lim }\limits_{x \to - 2} \frac{{(x + 2)\left( {{x^2} + 3x - 7} \right)}}{{(x + 2)(x - 9)}} = \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + 3x - 7}}{{x - 9}} = \frac{9}{{11}}$

a) Đúng

b) Đúng

c) Đúng

d) Sai

a) $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4 + x} - 2}}{{4x}} = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt {4 + x} - 2)(\sqrt {4 + x} + 2)}}{{4x(\sqrt {4 + x} + 2)}} = \mathop {\lim }\limits_{x \to 0} \frac{{4 + x - 4}}{{4x(\sqrt {4 + x} + 2)}}$

$= \mathop {\lim }\limits_{x \to 0} \frac{1}{{4(\sqrt {4 + x} + 2)}} = \frac{1}{{4(\sqrt 4 + 2)}} = \frac{1}{{16}}{\rm{. }}$

b)

$\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to 2} \frac{{4 - {x^2}}}{{\sqrt {x + 7} - 3}}}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - x)(2 + x)(\sqrt {x + 7} + 3)}}{{(\sqrt {x + 7} - 3)(\sqrt {x + 7} + 3)}} = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - x)(2 + x)(\sqrt {x + 7} + 3)}}{{x + 7 - 9}}}\\{}&{ = \mathop {\lim }\limits_{x \to 2} [ - (2 + x)(\sqrt {x + 7} + 3)] = - 4.6 = - 24}\end{array}$

c)

${\rm{ }}\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2x + 5} - 3}}{{\sqrt {x + 2} - 2}}}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(\sqrt {2x + 5} - 3)(\sqrt {2x + 5} + 3)(\sqrt {x + 2} + 2)}}{{(\sqrt {x + 2} - 2)(\sqrt {x + 2} + 2)(\sqrt {2x + 5} + 3)}}}\\{}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(2x + 5 - 9)(\sqrt {x + 2} + 2)}}{{(x + 2 - 4)(\sqrt {2x + 5} + 3)}} = \mathop {\lim }\limits_{x \to 2} \frac{{2(\sqrt {x + 2} + 2)}}{{\sqrt {2x + 5} + 3}} = \frac{4}{3}}\end{array}$

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

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

a) Đúng

b) Đúng

c) Sai

d) Đúng

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

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

c)

$\begin{array}{llllllllllllllllllll}{\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + x + 1} - 3x}}{{2 - 3x}}}&{ = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2}\left( {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} - 3x}}{{x\left( {\frac{2}{x} - 3} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3x}}{{x\left( {\frac{2}{x} - 3} \right)}}}\\{}&{ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3}}{{\frac{2}{x} - 3}} = \frac{{ - \sqrt 1 - 3}}{{ - 3}} = \frac{4}{3}}\end{array}$

d)

$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{8{x^3} + 3{x^2} + 1}} - x}}{{\sqrt {4{x^2} - x + 2} + 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{{x^3}\left( {8 + \frac{3}{x} + \frac{1}{{{x^3}}}} \right)}} - x}}{{\sqrt {{x^2}\left( {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} \right)} + 3x}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\sqrt[3]{{8 + \frac{3}{x} + \frac{1}{{{x^3}}}}} - x}}{{x\sqrt {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} + 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{8 + \frac{3}{x} + \frac{1}{{{x^3}}}}} - 1}}{{\sqrt {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} + 3}} = \frac{{\sqrt[3]{8} - 1}}{{ - \sqrt 4 + 3}} = 1\end{array}$

Ta có $g(x) = \frac{{2x + 1}}{{\sqrt {{x^2} + 2x} + \sqrt {{x^2} - 1} }} - 2m = \frac{{2 + \frac{1}{x}}}{{\sqrt {1 + \frac{2}{x}} + \sqrt {1 - \frac{1}{{{x^2}}}} }} - 2m$.

Do đó $\mathop {\lim }\limits_{x \to + \infty } g(x) = 1 - 2m = 0$. Vậy $m = \frac{1}{2}$.

Ta có $\mathop {\lim }\limits_{x \to - \infty } [(m - x)(mx + 1)] = \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( {\frac{m}{x} - 1} \right)\left( {m + \frac{1}{x}} \right) = - \infty$.

Do $\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{m}{x} - 1} \right)\left( {m + \frac{1}{x}} \right) = - m$ nên ta phải có $m > 0$.

Lấy dãy số $\left( {{x_n}} \right)$ bất kì sao cho ${x_n} \to + \infty$. Khi đó

$\left| {f\left( {{x_n}} \right)} \right| = \frac{{{{\sin }^2}{x_n}}}{{x_n^2}} \le \frac{1}{{x_n^2}} \to 0\,\,khi\,\,n \to + \infty .$

Vậy $\mathop {\lim }\limits_{n \to + \infty } f\left( {{x_n}} \right) = 0$. Từ đó $\mathop {\lim }\limits_{x \to + \infty } f(x) = 0$.

Đặt $f\left( x \right) = \left( {{m^2} + 1} \right){x^3} - 2{m^2}{x^2} - 4x + {m^2} + 1$.

Hàm số $f\left( x \right) = \left( {{m^2} + 1} \right){x^3} - 2{m^2}{x^2} - 4x + {m^2} + 1$ liên tục trên $\mathbb{R}$.

Ta có: $f\left( x \right) = {m^2}\left( {{x^3} - 2{x^2} + 1} \right) + {x^3} - 4x + 1$

$$f\left( { - 3} \right) = - 44{m^2} - 14 < 0;\,\,\forall m$$

$$f\left( 0 \right) = {m^2} + 1 > 0,\forall m\,$$

$$f\left( 1 \right) = - 2$$

$$f\left( 2 \right) = {m^2} + 1 > 0\,;\,\,\forall m$$

Vì $f\left( { - 3} \right).\,f\left( 0 \right) < 0$ nên phương trình có ít nhất 1 nghiệm thuộc khoảng $\left( { - 3;0} \right)$.

Vì $f\left( 0 \right).\,f\left( 1 \right) < 0$ nên phương trình có ít nhất 1 nghiệm thuộc khoảng $\left( {0;1} \right)$.

Vì $f\left( 1 \right).\,f\left( 2 \right) < 0$ nên phương trình có ít nhất 1 nghiệm thuộc khoảng $\left( {1;2} \right)$.

Vậy phương trình $\left( {{m^2} + 1} \right){x^3} - 2{m^2}{x^2} - 4x + {m^2} + 1 = 0$ có ít nhất 3 nghiệm trong khoảng $\left( { - 3;2} \right)$, mà phương trình đã cho là bậc 3 nên phương trình có đúng 3 nghiệm