Các dạng vô định của giới hạn

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: 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 $L$

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