Các dạng vô định

$\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } (x - 1)\sqrt {\dfrac{{{x^2}}}{{2{x^4} + {x^2} + 1}}}  \\= \mathop {\lim }\limits_{x \to  - \infty } \left[ { - \sqrt {\dfrac{{{x^2}{{(x - 1)}^2}}}{{2{x^4} + {x^2} + 1}}} } \right] \\= \mathop {\lim }\limits_{x \to  - \infty } \left[ { - \sqrt {\dfrac{{{x^2}({x^2} - 2x + 1)}}{{2{x^4} + {x^2} + 1}}} } \right]\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ { - \sqrt {\dfrac{{{x^4} - 2{x^3} + {x^2}}}{{2{x^4} + {x^2} + 1}}} } \right]\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ { - \sqrt {\dfrac{{1 - \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} } \right] =  - \dfrac{{\sqrt 2 }}{2}\end{array}$

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:

$= \dfrac{{1 + 0}}{{\sqrt {1 + 0 + 0}  + 1}} = \dfrac{1}{2}$

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

Bước 1:

\(\begin{array}{l}a{x^2} + bx - 5\\ = (ax + a + b)(x - 1) + a + b - 5\end{array}\)

Bước 2:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \left( {ax + a + b + \dfrac{{a + b - 5}}{{x - 1}}} \right) = 20\\ \Leftrightarrow \left\{ \begin{array}{l}a.1 + b + a = 20\\a + b - 5 = 0\end{array} \right.\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 15}\\{6 =  - 10}\end{array}} \right.\\ \Rightarrow P = {a^2} + {b^2} - a - b = 320\end{array}\)

$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)$.

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

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

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 $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x} = 1 + 1 + 1 = 3$

Đặt $x = \dfrac{1}{y}$, khi $x \to  + \infty :\,\,\,y \to 0$

$\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2)...(x + n)}} - x} \right) = \mathop {\lim }\limits_{y \to 0} \left( {\sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right)...\left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}} \right) = \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y}$

$\begin{array}{l}\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1\\ = \sqrt[n]{{1 + y}} - \sqrt[n]{{1 + y}} + \sqrt[n]{{\left( {1 + y} \right)\left( {1 + 2y} \right)}} - \sqrt[n]{{\left( {1 + y} \right)\left( {1 + 2y} \right)}} + ... + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}\\\,\,\,\,\, - \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}} + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1\\ = \left( {\sqrt[n]{{1 + y}} - 1} \right) + \sqrt[n]{{1 + y}}\left( {\sqrt[n]{{1 + 2y}} - 1} \right) + ... + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}\left( {\sqrt[n]{{1 + ny}} - 1} \right)\\ \Rightarrow \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y} = \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{\left( {\sqrt[n]{{1 + y}} - 1} \right)}}{y}} \right] + \mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{1 + y}}.\dfrac{{\left( {\sqrt[n]{{1 + 2y}} - 1} \right)}}{y}} \right] + ... + \\\,\,\,\,\,\mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}.\dfrac{{\left( {\sqrt[n]{{1 + ny}} - 1} \right)}}{y}} \right]\end{array}$

Tổng quát:

$\begin{array}{l}\mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}.\dfrac{{\sqrt[n]{{1 + ky}} - 1}}{y}} \right]\\ = \mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}.\dfrac{{\left( {\sqrt[n]{{1 + ky}} - 1} \right)\left[ {{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1} \right]}}{{y\left[ {{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1} \right]}}} \right]\\ = \mathop {\lim }\limits_{y \to 0} \dfrac{{(1 + ky - 1).\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}}}{{y{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1}}\\ = \mathop {\lim }\limits_{y \to 0} \dfrac{{k.\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}}}{{{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1}} = \dfrac{k}{n}\end{array}$

Khi đó:

$\mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y} = \dfrac{1}{n} + \dfrac{2}{n} + \dfrac{3}{n} + ... + \dfrac{n}{n} = \dfrac{{1 + 2 + 3 + ... + n}}{n} = \dfrac{{\dfrac{{n(n + 1)}}{2}}}{n} = \dfrac{{n + 1}}{2}$

Bước 1: Đưa \(\left| x \right|\) ra ngoài căn bậc hai: \(\sqrt {{x^2} + 3x + 5}  = \left| x \right|\sqrt {1 + \dfrac{3}{x} + \dfrac{5}{{{x^2}}}} \)

Bước 2: Phá dấu giá trị tuyệt đối và rút gọn x ở mẫu.

Bước 1:

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 20}}{{x - 2}}\) ta có \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = 10\) và \(f\left( x \right) - 20 = g\left( x \right)\left( {x - 2} \right) \Leftrightarrow f\left( x \right) = g\left( x \right)\left( {x - 2} \right) + 20\)

\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {g\left( x \right)\left( {x - 2} \right) + 20} \right] = 10.\left( {2 - 2} \right) + 20 = 20\)

Bước 2:

Ta có:

\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{6f\left( x \right) + 5 - 125}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 20}}{{x - 2}}.\dfrac{6}{{\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = 10.\dfrac{6}{{\left( {2 + 3} \right)\left[ {{{\left( {\sqrt[3]{{6.20 + 5}}} \right)}^2} + 5\sqrt[3]{{6.20 + 5}} + 25} \right]}} = \dfrac{4}{{25}}\end{array}\)

Bước 1: Tính \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 16}}{{x - 2}}\) ta có: \(f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 16\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {x - 2} \right)g\left( x \right) + 16} \right] = 16\).

Bước 2:

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {2f\left( x \right) - 16}  - 4}}{{{x^2} + x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) - 16 - 16}}{{\left( {{x^2} + x - 6} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) - 32}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 16}}{{x - 2}}.\mathop {\lim }\limits_{x \to 2} \dfrac{2}{{\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = 12.\dfrac{2}{{5.\left( {\sqrt {2.16 - 16}  + 4} \right)}} = \dfrac{3}{5}\end{array}\)

=> a=3; b=5

=> $a^2+b^2=34$

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{\sqrt {f\left( x \right) + 2}  - f\left( x \right)}}{{f\left( x \right) - 2}}\\\,\,\,\,\, = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right) + 2 - {f^2}\left( x \right)}}{{f\left( x \right) - 2}}.\dfrac{1}{{\sqrt {f\left( x \right) + 2}  + f\left( x \right)}}\\\,\,\,\,\, = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{ - \left[ {f\left( x \right) + 1} \right]\left[ {f\left( x \right) - 2} \right]}}{{f\left( x \right) - 2}}.\dfrac{1}{{\sqrt {f\left( x \right) + 2}  + f\left( x \right)}}\\\,\,\,\,\, =  - \dfrac{3}{4}\end{array}\)

Bước 1: Tính $\mathop {\lim }\limits_{x \to 4} f\left( x \right)$

Đặt \(\dfrac{{f\left( x \right) - 2018}}{{x - 4}} = g\left( x \right) \Rightarrow f\left( x \right) = \left( {x - 4} \right)g\left( x \right) + 2018\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 4} f\left( x \right) = 2018\).

Bước 2: Nhân cả tử và mẫu với $\sqrt{x}+2$. Tính $L$

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to 4} \dfrac{{1009\left[ {f\left( x \right) - 2018} \right]}}{{\left( {\sqrt x  - 2} \right)\left[ {\sqrt {2019f\left( x \right) + 2019}  + 2019} \right]}}\\\,\,\,\, = \mathop {\lim }\limits_{x \to 4} \dfrac{{1009\left[ {f\left( x \right) - 2018} \right]\left( {\sqrt x  + 2} \right)}}{{\left( {x - 4} \right)\left[ {\sqrt {2019f\left( x \right) + 2019}  + 2019} \right]}}\\\,\,\,\, = 1009.\mathop {\lim }\limits_{x \to 4} \dfrac{{f\left( x \right) - 2018}}{{x - 4}}.\dfrac{{\sqrt x  + 2}}{{\sqrt {2019f\left( x \right) + 2019}  + 2019}}\\\,\,\,\, = 1009.2019.\dfrac{{\sqrt {2018}  + 2}}{{\sqrt {2019.2018 + 2019}  + 2019}}\\\,\,\,\, = 1009.2019.\dfrac{{\sqrt 4  + 2}}{{2019 + 2019}} = 2018\end{array}\)