ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Giới hạn của hàm số

\[\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} + x} \right) = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right) = + \infty \]

vì\(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to + \infty } x = + \infty }\\{\mathop {\lim }\limits_{x \to + \infty } \sqrt {1 + \frac{1}{{{x^2}}}} + 1 = 2 >0}\end{array}} \right.\)

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\[\mathop {\lim }\limits_{x \to - \infty } {x^n} = - \infty \]

Bước 1:

Đặt \[g\left( x \right) = \frac{{f\left( x \right) - 20}}{{x - 2}}\] 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:

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