Cho n=2k+1,k∈N. Khi đó:
Ta có: \mathop {\lim }\limits_{x \to - \infty } {x^k} = + \infty nếu k chẵn và \mathop {\lim }\limits_{x \to - \infty } {x^k} = - \infty nếu k lẻ.
Do đó, vì n = 2k + 1,k \in N là số nguyên dương lẻ nên \mathop {\lim }\limits_{x \to - \infty } {x^n} = - \infty
Biết\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \dfrac{{a\sqrt 2 }}{b} + c với a, b, c \in \mathbb{Z} và \dfrac{a}{b} là phân số tối giản. Giá trị của a + b + c bằng:
Ta có \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - 2 + 2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}}
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - 2}}{{\sqrt 2 \left( {x - 1} \right)}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J.
Tính I = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - 2}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + x + 2 - 4}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2} + 2} \right)}}
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2} + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 2}}{{\sqrt 2 \left( {\sqrt {{x^2} + x + 2} + 2} \right)}} = \dfrac{3}{{4\sqrt 2 }}.
và J = \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{8 - 7x - 1}}{{\sqrt 2 \left( {x - 1} \right)\left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}}
\mathop { = \lim }\limits_{x \to 1} \dfrac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}} = \dfrac{{ - 7}}{{12\sqrt 2 }}.
Do đó \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2} - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J = \dfrac{{\sqrt 2 }}{{12}}
Suy ra a = 1, b = 12, c = 0. Vậy a + b + c = 13.
Cho hàm số f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\dfrac{{2x}}{{\sqrt {1 - x} }}}\,\,khi\,\,{x < 1}\\{\sqrt {3{x^2} + 1} }\,\,khi\,\,{x \ge 1}\end{array}} \right.. Khi đó \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) là:
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \sqrt {3{x^2} + 1} = \sqrt {{{3.1}^2} + 1} = 2
Khẳng định nào sau đây Sai?
\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} + 3x - 1} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( {1 + \dfrac{3}{x} - \dfrac{1}{{{x^2}}}} \right) = + \infty \end{array}
Cho đa thức f\left( x \right) thỏa mãn \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{x - 1}} = 12. Tính \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}
Đáp án: \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}
Đáp án: \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}
Bước 1:
Đặt g\left( x \right) = \dfrac{{f\left( x \right) - 2}}{{x - 1}} \Rightarrow f\left( x \right) = \left( {x - 1} \right)g\left( x \right) + 2
\Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {x - 1} \right)g\left( x \right) + 2} \right] = 2.
Bước 2:
Ta có:
\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{x - 1}}.\dfrac{1}{{\left( {x + 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = 12.\dfrac{1}{{2.\left( {2 + 1} \right)}} = 2\end{array}
