Giá trị của giới hạn $\,\mathop {\lim }\limits_{x \to + \infty } \,\,\left( {\sqrt {{x^2} + x} - \sqrt[3]{{{x^3} - {x^2}}}} \right)$ là:
\(\mathop {\lim }\limits_{x \to + \infty } \,\,\left( {\sqrt {{x^2} + x} - \sqrt[3]{{{x^3} - {x^2}}}} \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x} - x + x - \sqrt[3]{{{x^3} - {x^2}}}} \right)\)
\( = \mathop {\lim }\limits_{x \to + \infty } \left( {\dfrac{x}{{\sqrt {{x^2} + 1} + x}} + \dfrac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} - 1} \right)}^2}}}}}} \right) = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}.\)
Kết quả của giới hạn $\,\mathop {\lim }\limits_{x \to + \infty } x\sqrt {\dfrac{{2x + 1}}{{3{x^3} + {x^2} + 2}}} $ là:
$\mathop {\lim }\limits_{x \to + \infty } x\sqrt {\dfrac{{2x + 1}}{{3{x^3} + {x^2} + 2}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\dfrac{{{x^2}\left( {2x + 1} \right)}}{{3{x^3} + {x^2} + 2}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\dfrac{{2 + \dfrac{1}{x}}}{{3 + \dfrac{1}{x} + \dfrac{2}{{{x^3}}}}}} = \dfrac{{\sqrt 6 }}{3}.$
Tìm tất cả các giá trị của \(a\) để $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + 1} + ax} \right)$ là \( + \infty .\)
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 .$
Biết rằng \(L = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {4{x^2} - 2x + 1} + 2 - x}}{{\sqrt {a{x^2} - 3x} + bx}} > 0\) là hữu hạn (với \(a,\,b\) là tham số). Khẳng định nào dưới đây đúng.
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 \(\dfrac{{\sqrt {4{x^2} - 2x + 1} + 2 - x}}{{\sqrt {a{x^2} - 3x} + bx}} \sim \dfrac{{ - 3x}}{{\left( { - \sqrt a + b} \right)x}} = \dfrac{3}{{b - \sqrt a }} = L > 0 \Leftrightarrow b - \sqrt a > 0 \Rightarrow b > \sqrt a \)
Kết quả của giới hạn $\,\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {\dfrac{x}{{{x^2} - 1}}} $ là:
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$.
Tính \(\mathop {\lim }\limits_{x \to + \infty } \left( {x - \sqrt {{x^2} - 2x + 3} } \right)\)
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:
\( = \dfrac{{2 - 3.0}}{{1 + \sqrt {1 - 2.0 + 0} }} = 1\)
Giới hạn \(\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {x + 2} - \sqrt {8 - 2x} }}{{x - 2}}\) bằng:
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:
\( = \dfrac{3}{{\sqrt {2 + 2} + \sqrt {8 - 2.2} }} = \dfrac{3}{{2 + 2}} = \dfrac{3}{4}\)
Tìm giới hạn \(\mathop {\lim }\limits_{x \to - 3} \dfrac{{2x + 6}}{{{x^2} + x - 6}}\)
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:
\( = \dfrac{2}{{ - 3 - 2}} = \dfrac{{ - 2}}{5}\)
Tính $\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - x + 7} \right)$ bằng?
$\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - x + 7} \right) = {( - 1)^2} - ( - 1) + 7 = 9.$
Tính $\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} - 3x - 8} \right)$ bằng?
$\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} - 3x - 8} \right) = 3.{( - 2)^2} - 3.( - 2) - 8 = 12 + 6 - 8 = 10.$
Tính $\mathop {\lim }\limits_{x \to 2} \sqrt {\dfrac{{{x^4} + 3x - 1}}{{2{x^2} - 1}}} $ bằng?
$\mathop {\lim }\limits_{x \to 2} \sqrt {\dfrac{{{x^4} + 3x - 1}}{{2{x^2} - 1}}} = \sqrt {\dfrac{{{2^4} + 3.2 - 1}}{{{{2.2}^2} - 1}}} = \sqrt {\dfrac{{16 + 6 - 1}}{{8 - 1}}} = \sqrt 3 .$
Tính $\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^2} - 2x - 1}}{{{x^2} + 1}}$ bằng?
$\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^2} - 2x - 1}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3 - \dfrac{2}{x} - \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{{{x^2}}}}} = \dfrac{3}{1} = 3.$
Tính $\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}}$ bằng?
$\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{3x - 9}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{1}{3} = \dfrac{1}{3}.$
Trong các mệnh đề sau đâu là mệnh đề đúng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to - {1^ + }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} = \mathop {\lim }\limits_{x \to - {1^ + }} \dfrac{{(x + 1)(x + 2)}}{{x + 1}} \\= \mathop {\lim }\limits_{x \to - {1^ + }} (x + 2) = - 1 + 2 = 1\\\mathop {\lim }\limits_{x \to - {1^ - }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} = \mathop {\lim }\limits_{x \to - {1^ - }} \dfrac{{(x + 1)(x + 2)}}{{ - (x + 1)}} \\= \mathop {\lim }\limits_{x \to - {1^ - }} \left[ { - (x + 2)} \right] = - ( - 1 + 2) = - 1\\ \Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}} \ne \mathop {\lim }\limits_{x \to - {1^ - }} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}}\end{array}$
Suy ra, không tồn tại $\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 3x + 2}}{{\left| {x + 1} \right|}}.$
Tính $\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 4x + 3}}{{{x^2} - 9}}$ bằng?
$\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 4x + 3}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 1)(x - 3)}}{{(x - 3)(x + 3)}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{x - 1}}{{x + 3}} = \dfrac{{3 - 1}}{{3 + 3}} = \dfrac{1}{3}.$
Tính $\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 6x + 5}}{{{x^3} + 2{x^2} - 1}}$ bằng?
$\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 6x + 5}}{{{x^3} + 2{x^2} - 1}} = \mathop {\lim }\limits_{x \to - 1} \dfrac{{(x + 1)(x + 5)}}{{(x + 1)({x^2} + x - 1)}} = \mathop {\lim }\limits_{x \to - 1} \dfrac{{x + 5}}{{{x^2} + x - 1}} = \dfrac{{ - 1 + 5}}{{{{( - 1)}^2} + ( - 1) - 1}} = - 4$
Tính $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 4}}$ bằng?
\(\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 4}} \) \(= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 2)(x - 3)}}{{(x - 2)(x + 2)}} \) \(= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 3)}}{{x + 2}} \) \(= \dfrac{{(2 - 1)(2 - 3)}}{{2 + 2}} = \dfrac{{ - 1}}{4}\)
Tính $\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1} - 2}}{{\sqrt {3x} - 3}}$ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1} - 2}}{{\sqrt {3x} - 3}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(\sqrt {x + 1} - 2)(\sqrt {x + 1} + 2)(\sqrt {3x} + 3)}}{{(\sqrt {3x} - 3)(\sqrt {3x} + 3)(\sqrt {x + 1} + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(x + 1 - 4)(\sqrt {3x} + 3)}}{{(3x - 9)(\sqrt {x + 1} + 2)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)(\sqrt {3x} + 3)}}{{3(x - 3)(\sqrt {x + 1} + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {3x} + 3}}{{3(\sqrt {x + 1} + 2)}} \\= \dfrac{{\sqrt {3.3} + 3}}{{3(\sqrt {3 + 1} + 2)}} = \dfrac{1}{2}\end{array}$
Tính $\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}}$ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - \sqrt {x + 2} )(x + \sqrt {x + 2} )(\sqrt {4x + 1} + 3)}}{{(\sqrt {4x + 1} - 3)(\sqrt {4x + 1} + 3)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{({x^2} - x - 2)(\sqrt {4x + 1} + 3)}}{{(4x + 1 - 9)(x + \sqrt {x + 2} )}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(x - 2)(\sqrt {4x + 1} + 3)}}{{4(x - 2)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(\sqrt {4x + 1} + 3)}}{{4(x + \sqrt {x + 2} )}} \\= \dfrac{{(2 + 1)(\sqrt {4.2 + 1} + 3)}}{{4(2 + \sqrt {2 + 2} )}} = \dfrac{9}{8}\end{array}$
Tính $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}}$ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{(1 - \sqrt[3]{{x + 1}})\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - (x + 1)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{0 + 1}} + {{\left( {\sqrt[3]{{0 + 1}}} \right)}^2}} \right)}} = \dfrac{{ - 1}}{9}\end{array}$
