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?
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{(1 - \sqrt[3]{{x + 1}})\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - (x + 1)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{0 + 1}} + {{\left( {\sqrt[3]{{0 + 1}}} \right)}^2}} \right)}} = \dfrac{{ - 1}}{9}\end{array}$
Tính$\mathop {\lim }\limits_{x \to - \infty } (x - 1)\sqrt {\dfrac{{{x^2}}}{{2{x^4} + {x^2} + 1}}} $ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } (x - 1)\sqrt {\dfrac{{{x^2}}}{{2{x^4} + {x^2} + 1}}} \\= \mathop {\lim }\limits_{x \to - \infty } \left[ { - \sqrt {\dfrac{{{x^2}{{(x - 1)}^2}}}{{2{x^4} + {x^2} + 1}}} } \right] \\= \mathop {\lim }\limits_{x \to - \infty } \left[ { - \sqrt {\dfrac{{{x^2}({x^2} - 2x + 1)}}{{2{x^4} + {x^2} + 1}}} } \right]\\ = \mathop {\lim }\limits_{x \to - \infty } \left[ { - \sqrt {\dfrac{{{x^4} - 2{x^3} + {x^2}}}{{2{x^4} + {x^2} + 1}}} } \right]\\ = \mathop {\lim }\limits_{x \to - \infty } \left[ { - \sqrt {\dfrac{{1 - \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} } \right] = - \dfrac{{\sqrt 2 }}{2}\end{array}$
Tính $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 3} - x} \right)$ bằng?
Bước 1:
$\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 3} - x} \right) \\= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\sqrt {{x^2} + x + 3} - x} \right)\left( {\sqrt {{x^2} + x + 3} + x} \right)}}{{\left( {\sqrt {{x^2} + x + 3} + x} \right)}} \\= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + x + 3 - {x^2}}}{{\sqrt {{x^2} + x + 3} + x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x + 3}}{{\sqrt {{x^2} + x + 3} + x}} \end{array}$
Bước 2:
$= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{3}{x}}}{{\sqrt {1 + \dfrac{1}{x} + \dfrac{3}{{{x^2}}}} + 1}} $
Bước 3:
$= \dfrac{{1 + 0}}{{\sqrt {1 + 0 + 0} + 1}} = \dfrac{1}{2}$
Cho \(a,\,b\) là các số nguyên và \(\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}} = 20\). Tính \(P = {a^2} + {b^2} - a - b\)
Bước 1:
\(\begin{array}{l}a{x^2} + bx - 5\\ = (ax + a + b)(x - 1) + a + b - 5\end{array}\)
Bước 2:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \left( {ax + a + b + \dfrac{{a + b - 5}}{{x - 1}}} \right) = 20\\ \Leftrightarrow \left\{ \begin{array}{l}a.1 + b + a = 20\\a + b - 5 = 0\end{array} \right.\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 15}\\{6 = - 10}\end{array}} \right.\\ \Rightarrow P = {a^2} + {b^2} - a - b = 320\end{array}\)
Tính $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 1} + x - 1} \right)$ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 1} + x - 1} \right) \\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {{x^2} + 1} + x - 1} \right)\left( {\sqrt {{x^2} + 1} - x + 1} \right)}}{{\sqrt {{x^2} + 1} - x + 1}} \\= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2} + 1 - {{(x - 1)}^2}}}{{\sqrt {{x^2} + 1} - x + 1}}\\=\mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2} + 1 - {x^2} + 2x - 1}}{{\sqrt {{x^2} + 1} - x + 1}} \\= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x}}{{\sqrt {{x^2} + 1} - x + 1}} \\= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{{2x}}{x}}}{{\dfrac{{\sqrt {{x^2} + 1} }}{x} - \dfrac{x}{x} + \dfrac{1}{x}}} \\= \mathop {\lim }\limits_{x \to - \infty } \dfrac{2}{{ - \sqrt {1 + \dfrac{1}{{{x^2}}}} - 1 + \dfrac{1}{x}}}\\ = \dfrac{2}{{ - 1 - 1 + 0}} = - 1\end{array}$
Cho hàm số $f(x) = \sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} $. Khẳng định nào sau đây là đúng?
$f(x) = \sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} $
Ta có:
$\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } f(x) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}{{\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{({x^2} + 2x + 4) - ({x^2} - 2x + 4)}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{4}{{\sqrt {1 + \dfrac{2}{x} + \dfrac{4}{{{x^2}}}} + \sqrt {1 - \dfrac{2}{x} + \dfrac{4}{{{x^2}}}} }} = 2\end{array}$
$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}{{\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{({x^2} + 2x + 4) - ({x^2} - 2x + 4)}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{4x}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{{4x}}{x}}}{{\dfrac{{\sqrt {{x^2} + 2x + 4} }}{x} + \dfrac{{\sqrt {{x^2} - 2x + 4} }}{x}}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{4}{{ - \sqrt {1 + \dfrac{2}{x} + \dfrac{4}{{{x^2}}}} - \sqrt {1 - \dfrac{2}{x} + \dfrac{4}{{{x^2}}}} }} = \dfrac{4}{{ - 1 - 1}} = - 2\end{array}$
$ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } f(x) =- \mathop {\lim }\limits_{x \to - \infty } f(x)$.
Tính $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{{x^3} + 1}} + x - 1} \right)$ bằng?
\(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{{x^3} + 1}} + x - 1} \right)\) \( = \mathop {\lim }\limits_{x \to - \infty } \left( {x\sqrt[3]{{1 + \dfrac{1}{{{x^3}}}}} + x - 1} \right)\) \( = \mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( {\sqrt[3]{{1 + \dfrac{1}{{{x^3}}}}} + 1 - \dfrac{1}{x}} \right)} \right] = - \infty \)
Tính $\mathop {\lim }\limits_{x \to - \infty } x\sqrt {\dfrac{{3x + 2}}{{2{x^3} + {x^2} - 1}}} $ bằng?
$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } x\sqrt {\dfrac{{3x + 2}}{{2{x^3} + {x^2} - 1}}} = \mathop {\lim }\limits_{x \to - \infty } \left( { - \sqrt {\dfrac{{{x^2}\left( {3x + 2} \right)}}{{2{x^3} + {x^2} - 1}}} } \right) = \mathop {\lim }\limits_{x \to - \infty } \left( { - \sqrt {\dfrac{{3{x^3} + 2{x^2}}}{{2{x^3} + {x^2} - 1}}} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left( { - \sqrt {\dfrac{{3 + \dfrac{2}{x}}}{{2 + \dfrac{1}{x} - \dfrac{1}{{{x^3}}}}}} } \right) = - \sqrt {\dfrac{3}{2}} \end{array}$
Tính $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x}$
Ta có:
$\begin{array}{l}\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \sqrt {1 + 2x} - \sqrt {1 + 2x} + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} - \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \left( {\sqrt {1 + 2x} - 1} \right) + \sqrt {1 + 2x} \left( {\sqrt[3]{{1 + 3x}} - 1} \right) + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\end{array}$
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {1 + 2x} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right)\end{array}$
Tính:
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {1 + 2x} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {1 + 2x} - 1} \right)\left( {\sqrt {1 + 2x} + 1} \right)}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{\sqrt {1 + 2x} + 1}} = \dfrac{2}{{1 + 1}} = 1$
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\left( {\sqrt[3]{{1 + 3x}} - 1} \right)\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{3x}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{3\sqrt {1 + 2x} }}{{\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \dfrac{{3.1}}{{1 + 1 + 1}} = 1\end{array}$
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{4x}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}}}{{{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1}} = \dfrac{{4.1.1}}{{1 + 1 + 1 + 1}} = 1\end{array}$
Vậy $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x} = 1 + 1 + 1 = 3$
