Bài toán 3: Tìm $\mathop {\lim }\limits_{x \to \pm \infty } \frac{{f(x)}}{{g(x)}}$, trong đó $f(x),g(x) \to \infty $ (dạng vô định $\frac{\infty }{\infty }$). Phương pháp: Ta cần tìm cách đưa về các giới hạn: $\mathop {\lim }\limits_{x \to \pm \infty } {x^{2k}} = + \infty .$ $\mathop {\lim }\limits_{x \to + \infty } {x^{2k + 1}} = + \infty .$ $\mathop {\lim }\limits_{x \to – \infty } {x^{2k + 1}} = – \infty .$ $\mathop {\lim }\limits_{x \to \pm \infty } \frac{k}{{{x^n}}} = 0\left( {n > 0;k \ne 0} \right).$ $\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = \pm \infty $ $ \Leftrightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{k}{{f\left( x \right)}} = 0\left( {k \ne 0} \right).$ Ví dụ 9. Tìm các giới hạn sau: 1. $A = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{(4x + 1)}^3}{{(2x + 1)}^4}}}{{{{(3 + 2x)}^7}}}.$ 2. $B = \mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {4{x^2} – 3x + 4} + 3x}}{{\sqrt {{x^2} + x + 1} – x}}.$ 1. Ta có: $A = $ $\mathop {\lim }\limits_{x \to + \infty } \frac{{{{\left( {4 + \frac{1}{x}} \right)}^3}{{\left( {2 + \frac{1}{x}} \right)}^4}}}{{{{\left( {\frac{3}{x} + 2} \right)}^7}}}$ $ = 8.$ 2. Ta có: $B = $ $\mathop {\lim }\limits_{x \to – \infty } \frac{{ – \sqrt {4 – \frac{3}{x} + \frac{4}{{{x^2}}}} + 3}}{{ – \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 1}}$ $ = \frac{1}{2}.$ Ví dụ 10. Tìm các giới hạn sau: 1. $A = $ $\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {2{x^2} + 1} – \sqrt {{x^2} + 1} }}{{2x + 2}}.$ 2. $B = $ $\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {3{x^2} – 2} + \sqrt {x + 1} }}{{\sqrt {{x^2} + 1} – 1}}.$ 1. Ta có: $A = $ $\mathop {\lim }\limits_{x \to + \infty } \frac{{\left| x \right|\sqrt {2 + \frac{1}{{{x^2}}}} – \left| x \right|\sqrt {1 + \frac{1}{{{x^2}}}} }}{{x(2 + \frac{2}{x})}}$ $ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {2 + \frac{1}{{{x^2}}}} – \sqrt {1 + \frac{1}{{{x^2}}}} }}{{2 + \frac{2}{x}}}$ $ = \frac{{\sqrt 2 – 1}}{2}.$ 2. Ta có: $B = $ $\mathop {\lim }\limits_{x \to – \infty } \frac{{\left| x \right|\sqrt {3 – \frac{2}{{{x^2}}}} + \left| x \right|\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} }}{{\left| x \right|\left( {\sqrt {1 + \frac{1}{{{x^2}}}} – \frac{1}{{\left| x \right|}}} \right)}}$ $ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – \sqrt {3 – \frac{2}{{{x^2}}}} – \sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} }}{{ – \left( {\sqrt {1 + \frac{1}{{{x^2}}}} – \frac{1}{{\left| x \right|}}} \right)}}$ $ = \sqrt 3 .$
