Các dạng vô định

Bước 1:

$\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \left( {x + 2} \right)\end{array}$

Bước 2:

$= 2 + 2 = 4$

Bước 1:

$\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 2x}  - x} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{{\left( {\sqrt {{x^2} + 2x} } \right)}^2} - {x^2}}}{{\sqrt {{x^2} + 2x}  + x}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x}}{{\sqrt {{x^2} + 2x}  + x}}\end{array}$

Bước 2:

$\begin{array}{l} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x}}{{x\left( {\sqrt {1 + \dfrac{2}{x}}  + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{2}{{\sqrt {1 + \dfrac{2}{x}}  + 1}}\\ = \dfrac{2}{{1 + 1}} = 1\end{array}$

$\mathop {\lim }\limits_{x \to 1} \dfrac{{x - {x^3}}}{{\left( {2x - 1} \right)\left( {{x^4} - 3} \right)}} = \dfrac{{1 - {1^3}}}{{\left( {2.1 - 1} \right)\left( {{1^4} - 3} \right)}} = 0$

Ta có: $\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{3{x^2} - 4}} - \sqrt {3x - 2} }}{{x + 1}} = \dfrac{{\sqrt[3]{{12 - 4}} - \sqrt {6 - 2} }}{3} = \dfrac{0}{3} = 0$

Ta có $\mathop {\lim }\limits_{x \to  - 1} \dfrac{{\sqrt {3{x^2} + 1}  - x}}{{x - 1}} = \dfrac{{\sqrt {3 + 1}  + 1}}{{ - 1 - 1}} =  - \dfrac{3}{2}$

Ta có $\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2{x^2} + 5x - 3}}{{{x^2} + 6x + 3}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2 + \dfrac{5}{x} - \dfrac{3}{{{x^2}}}}}{{1 + \dfrac{6}{x} + \dfrac{3}{{{x^2}}}}} = 2$.

$\mathop {\lim }\limits_{x \to 1} \left( { - {x^2} + 3x - 2} \right) =  - {1^2} + 3.1 - 2 = 0$

Ta có $\left| {x + 2} \right| = x + 2$ với mọi $x >  - 2,$ do đó :

$\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{\left| {3x + 6} \right|}}{{x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{3\left| {x + 2} \right|}}{{x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{3\left( {x + 2} \right)}}{{x + 2}} = 3$

$\mathop {\lim }\limits_{x \to  - 3} \left| {\dfrac{{ - {x^2} - x + 6}}{{{x^2} + 3x}}} \right| = \mathop {\lim }\limits_{x \to  - 3} \left| {\dfrac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{x\left( {x + 3} \right)}}} \right|$ $= \mathop {\lim }\limits_{x \to  - 3} \left| {\dfrac{{x - 2}}{x}} \right| = \left| {\dfrac{{ - 3 - 2}}{{ - 3}}} \right| = \dfrac{5}{3}$

Ta có $\mathop {\lim }\limits_{x \to  - \sqrt 3 } \dfrac{{2({x^3} + 3\sqrt 3 )}}{{3 - {x^2}}}$ $= \mathop {\lim }\limits_{x \to  - \sqrt 3 } \dfrac{{2\left( {x + \sqrt 3 } \right)\left( {{x^2} - \sqrt 3 x + 3} \right)}}{{\left( {\sqrt 3  - x} \right)\left( {\sqrt 3  + x} \right)}}$ $= \mathop {\lim }\limits_{x \to  - \sqrt 3 } \dfrac{{2\left( {{x^2} - \sqrt 3 x + 3} \right)}}{{\sqrt 3  - x}}$

$= \dfrac{{2\left[ {{{\left( { - \sqrt 3 } \right)}^2} - \sqrt 3 .\left( { - \sqrt 3 } \right) + 3} \right]}}{{\sqrt 3  - \left( { - \sqrt 3 } \right)}} = \dfrac{{18}}{{2\sqrt 3 }} = 3\sqrt 3$ $\Rightarrow \left\{ \begin{array}{l}a = 3\\b = 0\end{array} \right. \Rightarrow {a^2} + {b^2} = 9$.

Ta có $3 - x > 0$ với mọi $x < 3,$ do đó:

$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{3 - x}}{{\sqrt {27 - {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{3 - x}}{{\sqrt {\left( {3 - x} \right)\left( {9 + 3x + {x^2}} \right)} }}$

$= \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{\sqrt {3 - x} }}{{\sqrt {9 + 3x + {x^2}} }} = \dfrac{{\sqrt {3 - 3} }}{{\sqrt {9 + 3.3 + {3^2}} }} = 0.$

$\mathop {\lim }\limits_{x \to  - 1} \,\,\dfrac{{{x^5} + 1}}{{{x^3} + 1}} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{\left( {x + 1} \right)\left( {{x^4} - {x^3} + {x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}$ $= \mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^4} - {x^3} + {x^2} - x + 1}}{{{x^2} - x + 1}} = \dfrac{5}{3}$

Ta có $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {{x^2} + x}  - \sqrt x }}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left( {{x^2} + x} \right) - x}}{{{x^2}\left( {\sqrt {{x^2} + x}  + \sqrt x } \right)}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{\sqrt {{x^2} + x}  + \sqrt x }} =  + \infty$

vì $1 > 0$; $\mathop {\lim }\limits_{x \to {0^ + }} \left( {\sqrt {{x^2} + x}  + \sqrt x } \right) = 0$ và $\sqrt {{x^2} + x}  + \sqrt x  > 0$ với mọi $x > 0.$

Ta có $\mathop {\lim }\limits_{x \to 1} \,\,\dfrac{{\sqrt[3]{x} - 1}}{{\sqrt[3]{{4x + 4}} - 2}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{(x - 1)\left( {\sqrt[3]{{{{\left( {4x + 4} \right)}^2}}} + 2\sqrt[3]{{4x + 4}} + 4} \right)}}{{\left( {4x + 4 - 8} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}$

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

Ta có $\mathop {\lim }\limits_{x \to 0} \dfrac{{2\sqrt {1 + x}  - \sqrt[3]{{8 - x}}}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{2\sqrt {1 + x}  - 2}}{x} + \dfrac{{2 - \sqrt[3]{{8 - x}}}}{x}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{2}{{\sqrt {x + 1}  + 1}} + \dfrac{1}{{4 + 2\sqrt[3]{{8 - x}} + \sqrt[3]{{{{\left( {8 - x} \right)}^2}}}}}} \right) = 1 + \dfrac{1}{{12}} = \dfrac{{13}}{{12}}.$

$\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {4{x^2} - 2x + 1}  + 2 - x}}{{\sqrt {9{x^2} - 3x}  + 2x}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {4 - \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}  + \dfrac{2}{x} - 1}}{{\sqrt {9 - \dfrac{3}{x}}  + 2}} = \dfrac{1}{5}.$

Bước 1:

Ta có $\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {1 + 2{x^2}}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } x\left( {\sqrt {\dfrac{1}{{{x^2}}} + 2}  - 1} \right)$

Bước 2:

Vì $\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty$ và $\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {\dfrac{1}{{{x^2}}} + 2}  - 1} \right) = \sqrt 2  - 1 > 0.$

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

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

Bước 1:

Ta có $\mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\dfrac{a}{{1 - x}} - \dfrac{b}{{1 - {x^3}}}} \right)$ $= \mathop {\lim }\limits_{x \to 1} \dfrac{{a + ax + a{x^2} - b}}{{1 - {x^3}}}$ $= \mathop {\lim }\limits_{x \to 1} \dfrac{{a + ax + a{x^2} - b}}{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}.$

Bước 2:

Khi đó $\mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\dfrac{a}{{1 - x}} - \dfrac{b}{{1 - {x^3}}}} \right)$ hữu hạn thì tử thức $a + ax + a{x^2} - b$ phải có nghiệm bằng $1$

$\Leftrightarrow a + a.1 + a{.1^2} - b = 0 \Leftrightarrow 3a - b =  0.$

Vậy ta có $\left\{ \begin{array}{l}a + b = 4\\3a - b =  0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 3\end{array} \right.$

Bước 3:

$\Rightarrow L =  - \mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\dfrac{a}{{1 - x}} - \dfrac{b}{{1 - {x^3}}}} \right)$

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

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

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