Bài tập Giới hạn của hàm số tại một điểm và tại vô cực lớp 11 (có lời giải)

Đáp án đúng là: A

 $\underset{x\to 3}{\lim }\left(x^2+x+5\right)=3^2+3+5=17$.

Đáp án đúng là: C

 $\underset{x\to 2}{\lim }\frac{3x-5}{{\left(x-2\right)}^2}$ = +∞ vì  $\left\{\begin{array}{c}\begin{array}{c}\underset{x\to 2}{\lim }\left(3x-5\right)=1>0\\ \underset{x\to 2}{\lim }{\left(x-2\right)}^2=0\end{array}\\ \left(x-2\right)>0\forall x\ne 2\end{array}\right.$

Đáp án đúng là: C

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

Vì  $\left\{\begin{array}{c}\underset{x\to \infty }{\lim }{x}^4=+\infty \\ \underset{x\to \infty }{\lim }\left(1+\frac{2}{x^2}+\frac{3}{x^4}\right)=1>0\end{array}\right.$

Đáp án đúng là: D

 $\underset{x\to 2}{\lim }\frac{x^2-5x+6}{x^2-6x+8}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x-4\right)}=\underset{x\to 2}{\lim }\frac{x-3}{x-4}=\frac{1}{2}$.

Đáp án đúng là: D

 $\underset{x\to 1}{\lim }\frac{\sqrt[3]{x}-1}{x-1}=\underset{x\to 1}{\lim }\frac{\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$

 $=\underset{x\to 1}{\lim }\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\frac{1}{3}$.

Đáp án đúng là: A

 $\underset{x\to -\infty }{\lim }\left(\sqrt{3-x^2}-\sqrt{7-x^2}\right)=\underset{x\to -\infty }{\lim }\frac{3-x^2-\left(7-x^2\right)}{\sqrt{3-x^2}+\sqrt{7-x^2}}$

 $=\underset{x\to -\infty }{\lim }\frac{-4}{\sqrt{3-x^2}+\sqrt{7-x^2}}=0$.

Hướng dẫn giải:

Đáp án đúng là: C

 $\underset{x\to 0}{\lim }\left(\frac{2}{x}-\frac{3}{x^2}\right)=\underset{x\to 0}{\lim }\frac{2x-3}{x^2}=-\infty$ vì  $\left\{\begin{array}{c}\underset{x\to 0}{\lim }\left(2x-3\right)=-3\\ \underset{x\to 0}{\lim }{x}^2=0\\ x^2>0\forall x\ne 0\end{array}\right.$

Đáp án đúng là: C

 $\underset{x\to \infty }{\lim }\frac{\sqrt{3x^2-5}-\sqrt{2x^2+1}}{\sqrt{5x^2-7}+2}=\underset{x\to \infty }{\lim }\frac{\left|x\right|\sqrt{3-\frac{5}{x^2}}-\left|x\right|\sqrt{2+\frac{1}{x^2}}}{\left|x\right|\sqrt{5-\frac{7}{x^2}}+2}$

$=\underset{x\to \infty }{\lim }\frac{-x\sqrt{3-\frac{5}{x^2}}+x\sqrt{2+\frac{1}{x^2}}}{-x\sqrt{5-\frac{7}{x^2}}+2}=\underset{x\to \infty }{\lim }\frac{\sqrt{3-\frac{5}{x^2}}-\sqrt{2+\frac{1}{x^2}}}{\sqrt{5-\frac{7}{x^2}}-\frac{2}{x}}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{15}}{5}$

Hướng dẫn giải:

Đáp án đúng là: A

Ta có: \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 7x + 11} \right) = 3 \cdot {2^2} + 7 \cdot 2 + 11 = 37\).

Hướng dẫn giải:

Đáp án đúng là: D

Ta có: \(H = \mathop {\lim }\limits_{x \to 1} \frac{{1 - x}}{{\sqrt {x + 1} }} = \frac{{1 - 1}}{{\sqrt {1 + 1} }} = 0\) và \(K = \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} + {x^2} + 2}}{{{x^3} + x + 3}} = \frac{{{1^4} + {1^2} + 2}}{{{1^3} + 1 + 3}} = \frac{4}{5}\)

Do đó, H + K = 0 + \(\frac{4}{5}\)=\(\frac{4}{5}\).

Hướng dẫn giải:

Đáp án đúng là: B

Ta có: \(\mathop {\lim }\limits_{x \to 2} \left( {\sqrt {2{x^2} + 1} + ax} \right) = \sqrt {2 \cdot {2^2} + 1} + 2a = 3 + 2a = 0 \Leftrightarrow a = - \frac{3}{2}\).

Do đó, a =\( - \frac{3}{2}\) để \(\mathop {\lim }\limits_{x \to 2} \left( {\sqrt {2{x^2} + 1} + ax} \right)\) bằng 0.

Hướng dẫn giải:

Đáp án đúng là: D

Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^2} + 5x - 3}}{{{x^2} + 6x + 3}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 + \frac{5}{x} - \frac{3}{{{x^2}}}}}{{1 + \frac{6}{x} + \frac{3}{{{x^2}}}}} = 2\).

Hướng dẫn giải:

Đáp án đúng là: D

Ta có:

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

Vậy a = 3.

Hướng dẫn giải:

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

a) Sai. Ta có: \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3{x^2} + 1} }}{{x + 1}} = \frac{{\sqrt {3 \cdot {1^2} + 1} }}{{1 + 1}} = 1\)

\(\mathop {\lim }\limits_{x \to - 2} g\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 2} \frac{{2 \cdot \left( { - 2} \right)}}{{ - 2 + 1}} = 4\).

b) Đúng. Ta có: \(\mathop {\lim }\limits_{x \to + \infty } g\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{2}{{1 + \frac{1}{x}}} = 2\).

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

c) Sai. Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) - \mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = - \sqrt 3 - 2\).

d) Đúng. Ta có:

\(\mathop {\lim }\limits_{x \to - 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {3{x^2} + 1} + 2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {\sqrt {3{x^2} + 1} + 2x} \right) \cdot \left( {\sqrt {3{x^2} + 1} - 2x} \right)}}{{\left( {x + 1} \right)\left( {\sqrt {3{x^2} + 1} - 2x} \right)}}\)

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

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

Hướng dẫn giải:

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

a) Đúng. Ta có: \[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 3}}{{1 - x}} = \frac{{{2^2} - 4 \cdot 2 + 3}}{{1 - 2}} = 1\].

b) Sai. Ta có: \[\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{1 - x}} = \frac{{{3^2} - 4 \cdot 3 + 3}}{{1 - 3}} = 0\].

c) Đúng. Ta có: \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{1 - x}} = \mathop {\lim }\limits_{x \to 1} \left( {3 - x} \right) = 3 - 1 = 2\].

d) Sai. Ta có:

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

Hướng dẫn giải:

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

\[L = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + ax + 5} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{ax + 5}}{{\sqrt {{x^2} + ax + 5} - x}}\]

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

a) Đúng. L = 5 $\iff$\( - \frac{a}{2} = 5 \Leftrightarrow a = - 10\).

b) Sai. L > 0 $\iff$\( - \frac{a}{2} > 0 \Leftrightarrow a < 0\).

c) Đúng. L < 0 $\iff$\( - \frac{a}{2} < 0 \Leftrightarrow a > 0\).

d) Đúng. L =-1 $\iff$\( - \frac{a}{2} = - 1 \Leftrightarrow a = 2\)

x² – 3x + 2 = 0 $\iff$\(\left[ \begin{array}{l}x = 1\\x = 2\end{array} \right.\).

Hướng dẫn giải:

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

Ta có: f(x) = x² – 3x + 2 = (x – 2)(x – 1).

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

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

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

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

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

Vậy a + 3b =\( - \frac{1}{2} + 3 \cdot \frac{1}{2} = 1\).

Hướng dẫn giải

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

Ta có: \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {a{x^2} + bx} - cx} \right) = - 2 \Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {a - {c^2}} \right){x^2} + bx}}{{\sqrt {a{x^2} + bx} + cx}} = - 2\)

Điều này xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}a - {c^2} = 0\,\,\,\left( {a,c > 0} \right)\\\frac{b}{{\sqrt a + c}} = - 2\end{array} \right.\).

Vì nếu (c $\le$ 0 thì \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {a{x^2} + bx} - cx} \right) = + \infty \)).

Mặt khác, ta lại có: c² + a = 18.

Do đó, \(\left\{ \begin{array}{l}a = {c^2} = 9\\b = - 2\left( {\sqrt a + c} \right)\end{array} \right.\)$\iff$ a = 9, b =-12, c = 3.

a) Đúng. a = 9

b) Sai. Ta có: b =-4c.

c) Đúng. Ta có: a = 3c.

d) Sai. P = a + b + 5c = 9 + (-12) + 5.3 = 12.

Hướng dẫn giải:

Đáp án: 2025.

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

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

\( \Leftrightarrow \left\{ \begin{array}{l}a + 1 = 0\\a + b + 3 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 1\\b = - 1\end{array} \right.\)$\Rightarrow$ T = 2024a – 4049b = 2024.(-1) – 4049.(-1) = 2025.

Hướng dẫn giải:

Đáp án: 0,75.

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

\( = \mathop {\lim }\limits_{x \to 1} \frac{3}{{\left( {2x - 1} \right)\left( {\sqrt {3x + 1} + 2} \right)}} = \frac{3}{4} = 0,75\).

Hướng dẫn giải:

Đáp án: 2032.

Ta có: \(\frac{{a\sqrt b }}{c} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2023x + 2}}{{\sqrt {3{x^2} + 2023x + 2} + x\sqrt 3 }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2023 + \frac{2}{x}}}{{\sqrt {3 + \frac{{2023}}{x} + \frac{2}{{{x^2}}}} + \sqrt 3 }} = \frac{{2023\sqrt 3 }}{6}\)

Vậy a + b + c = 2023 + 3 + 6 = 2032.

Hướng dẫn giải:

Đáp án: 2.

\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x} - \sqrt[3]{{{x^3} + 1}}} \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x} - x} \right) + \mathop {\lim }\limits_{x \to + \infty } \left( {x - \sqrt[3]{{{x^3} + 1}}} \right)\)

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

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

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

Xét B = \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1}}{{{x^2} + x\sqrt[3]{{{x^3} + 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}}} = 0\)

(Do \(\mathop {\lim }\limits_{x \to + \infty } \left( {{x^2} + x\sqrt[3]{{{x^3} + 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}} \right) = + \infty \))

Suy ra \[\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x} - \sqrt[3]{{{x^3} + 1}}} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x}}{{x\sqrt {1 - \frac{1}{x}} + x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x}}{{x\left( {\sqrt {1 - \frac{1}{x}} + 1} \right)}} \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 1\end{array} \right.\]

Do đó, P = a + b² = 1 + 1² = 2.

Hướng dẫn giải:

Đáp án: -1.

Ta có: \(\mathop {\lim }\limits_{x \to - 3} \frac{{f\left( x \right) - 2}}{{x + 2}} = 5 \Leftrightarrow \frac{{\mathop {\lim }\limits_{x \to - 3} \left( {f\left( x \right) - 2} \right)}}{{\mathop {\lim }\limits_{x \to - 3} \left( {x + 2} \right)}} = 5\)

\( \Leftrightarrow \frac{{\mathop {\lim }\limits_{x \to - 3} \left( {f\left( x \right) - 2} \right)}}{{ - 1}} = 5 \Leftrightarrow \mathop {\lim }\limits_{x \to - 3} f\left( x \right) = - 3\).

Vậy \(\mathop {\lim }\limits_{x \to - 3} \frac{{ - 4 + \sqrt[3]{{5f\left( x \right) + 7}}}}{{{x^2} + x}} = \frac{{ - 4 + \sqrt[3]{{5 \cdot \left( { - 3} \right) + 7}}}}{{{{\left( { - 3} \right)}^2} - 3}} = - 1\)