Cho \(n = 2k + 1,k \in N\). 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:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{x - 1}}.\dfrac{1}{{\left( {x + 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = 12.\dfrac{1}{{2.\left( {2 + 1} \right)}} = 2\end{array}\)
