Các dạng vô định

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

$= \mathop {\lim }\limits_{x \to  + \infty } \left( {\dfrac{x}{{\sqrt {{x^2} + 1}  + x}} + \dfrac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} - 1} \right)}^2}}}}}} \right) = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}.$

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

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

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

Ta phải có .$a{x^2} - 3x > 0$. trên $\left( { - \infty ;\alpha } \right) \Leftrightarrow a \ge 0$

Ta có $x \to  - \infty  \Rightarrow \sqrt {4{x^2} - 2x + 1}  + 2 - x \sim \sqrt {4{x^2}}  - x =  - 3x \ne 0$

Như vậy xem như “tử” là một đa thức bậc 1. Khi đó $\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {4{x^2} - 2x + 1}  + 2 - x}}{{\sqrt {a{x^2} - 3x}  + bx}} > 0$ khi và chỉ khi $\sqrt {a{x^2} - 3x}  + bx$ là đa thức bậc 1.

Ta có

Khi đó $\dfrac{{\sqrt {4{x^2} - 2x + 1}  + 2 - x}}{{\sqrt {a{x^2} - 3x}  + bx}} \sim \dfrac{{ - 3x}}{{\left( { - \sqrt a  + b} \right)x}} = \dfrac{3}{{b - \sqrt a }} = L > 0 \Leftrightarrow b - \sqrt a  > 0 \Rightarrow b > \sqrt a$

Với $x \in \left( { - 1;0} \right)$ thì $x + 1 > 0$ và $\dfrac{x}{{x - 1}} > 0$.

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

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

Bước 1:

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

Bước 2:

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

Bước 3:

$= \dfrac{{2 - 3.0}}{{1 + \sqrt {1 - 2.0 + 0} }} = 1$

Bước 1:

Ta có:

$\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {x + 2}  - \sqrt {8 - 2x} }}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 2 - 8 + 2x}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + \sqrt {8 - 2x} } \right)}}\end{array}$

Bước 2:

$\begin{array}{l} = \mathop {\lim }\limits_{x \to 2} \dfrac{{3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + \sqrt {8 - 2x} } \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{3}{{\sqrt {x + 2}  + \sqrt {8 - 2x} }}\end{array}$

Bước 3:

$= \dfrac{3}{{\sqrt {2 + 2}  + \sqrt {8 - 2.2} }} = \dfrac{3}{{2 + 2}} = \dfrac{3}{4}$

Bước 1:

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

Bước 2:

$= \dfrac{2}{{ - 3 - 2}} = \dfrac{{ - 2}}{5}$

$\mathop {\lim }\limits_{x \to  - 1} \left( {{x^2} - x + 7} \right) = {( - 1)^2} - ( - 1) + 7 = 9.$

$\mathop {\lim }\limits_{x \to  - 2} \left( {3{x^2} - 3x - 8} \right) = 3.{( - 2)^2} - 3.( - 2) - 8 = 12 + 6 - 8 = 10.$

$\mathop {\lim }\limits_{x \to 2} \sqrt {\dfrac{{{x^4} + 3x - 1}}{{2{x^2} - 1}}}  = \sqrt {\dfrac{{{2^4} + 3.2 - 1}}{{{{2.2}^2} - 1}}}  = \sqrt {\dfrac{{16 + 6 - 1}}{{8 - 1}}}  = \sqrt 3 .$

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

$\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{3x - 9}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{1}{3} = \dfrac{1}{3}.$

$\begin{array}{l}\mathop {\lim }\limits_{x \to  - {1^ + }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} = \mathop {\lim }\limits_{x \to  - {1^ + }} \dfrac{{(x + 1)(x + 2)}}{{x + 1}} \\= \mathop {\lim }\limits_{x \to  - {1^ + }} (x + 2) =  - 1 + 2 = 1\\\mathop {\lim }\limits_{x \to  - {1^ - }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} = \mathop {\lim }\limits_{x \to  - {1^ - }} \dfrac{{(x + 1)(x + 2)}}{{ - (x + 1)}} \\= \mathop {\lim }\limits_{x \to  - {1^ - }} \left[ { - (x + 2)} \right] =  - ( - 1 + 2) =  - 1\\ \Rightarrow \mathop {\lim }\limits_{x \to  - {1^ + }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} \ne \mathop {\lim }\limits_{x \to  - {1^ - }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}}\end{array}$

Suy ra, không tồn tại $\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}}.$

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

$\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 6x + 5}}{{{x^3} + 2{x^2} - 1}} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{(x + 1)(x + 5)}}{{(x + 1)({x^2} + x - 1)}} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{x + 5}}{{{x^2} + x - 1}} = \dfrac{{ - 1 + 5}}{{{{( - 1)}^2} + ( - 1) - 1}} =  - 4$

$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 4}}$ $= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 2)(x - 3)}}{{(x - 2)(x + 2)}}$ $= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 3)}}{{x + 2}}$ $= \dfrac{{(2 - 1)(2 - 3)}}{{2 + 2}} = \dfrac{{ - 1}}{4}$

$\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1}  - 2}}{{\sqrt {3x}  - 3}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(\sqrt {x + 1}  - 2)(\sqrt {x + 1}  + 2)(\sqrt {3x}  + 3)}}{{(\sqrt {3x}  - 3)(\sqrt {3x}  + 3)(\sqrt {x + 1}  + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(x + 1 - 4)(\sqrt {3x}  + 3)}}{{(3x - 9)(\sqrt {x + 1}  + 2)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)(\sqrt {3x}  + 3)}}{{3(x - 3)(\sqrt {x + 1}  + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {3x}  + 3}}{{3(\sqrt {x + 1}  + 2)}} \\= \dfrac{{\sqrt {3.3}  + 3}}{{3(\sqrt {3 + 1}  + 2)}} = \dfrac{1}{2}\end{array}$

$\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1}  - 3}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - \sqrt {x + 2} )(x + \sqrt {x + 2} )(\sqrt {4x + 1}  + 3)}}{{(\sqrt {4x + 1}  - 3)(\sqrt {4x + 1}  + 3)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{({x^2} - x - 2)(\sqrt {4x + 1}  + 3)}}{{(4x + 1 - 9)(x + \sqrt {x + 2} )}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(x - 2)(\sqrt {4x + 1}  + 3)}}{{4(x - 2)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(\sqrt {4x + 1}  + 3)}}{{4(x + \sqrt {x + 2} )}} \\= \dfrac{{(2 + 1)(\sqrt {4.2 + 1}  + 3)}}{{4(2 + \sqrt {2 + 2} )}} = \dfrac{9}{8}\end{array}$

$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{(1 - \sqrt[3]{{x + 1}})\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - (x + 1)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{0 + 1}} + {{\left( {\sqrt[3]{{0 + 1}}} \right)}^2}} \right)}} = \dfrac{{ - 1}}{9}\end{array}$