Phương pháp: Ta sử dụng các công thức lượng giác biến đổi về các dạng sau: + $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}}$ $ = 1$, từ đó suy ra $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}}$ $ = 1.$ + Nếu $\mathop {\lim }\limits_{x \to {x_0}} u(x) = 0$ $ \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin u(x)}}{{u(x)}} = 1$ và $\mathop {\lim }\limits_{x \to {x_0}} \frac{{\tan u(x)}}{{u(x)}} = 1.$ Ví dụ 13. Tìm các giới hạn sau: 1. $A = $ $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos x} – \sqrt[3]{{\cos x}}}}{{{{\sin }^2}x}}.$ 2. $B = $ $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} – \sqrt[3]{{1 + 3x}}}}{{1 – \sqrt {\cos 2x} }}.$ 1. Ta có: $A = $ $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos x} – 1}}{{{x^2}}}\frac{{{x^2}}}{{{{\sin }^2}x}}$ $ + \mathop {\lim }\limits_{x \to 0} \frac{{1 – \sqrt[3]{{\cos x}}}}{{{x^2}}}.\frac{{{x^2}}}{{{{\sin }^2}x}}.$ Mà: $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos x} – 1}}{{{x^2}}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x – 1}}{{{x^2}}}.\frac{1}{{\sqrt {\cos x} + 1}}$ $ = – \frac{1}{4}.$ $\mathop {\lim }\limits_{x \to 0} \frac{{1 – \sqrt[3]{{\cos x}}}}{{{x^2}}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{{x^2}}}$$.\frac{1}{{\sqrt[3]{{{{\cos }^2}x}} + \sqrt[3]{{\cos x}} + 1}}$ $ = \frac{1}{6}.$ Do đó: $A = – \frac{1}{4} + \frac{1}{6} = – \frac{1}{{12}}.$ 2. Ta có: $B = $ $\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt {1 + 2x} – \sqrt[3]{{1 + 3x}}}}{{{x^2}}}}}{{\frac{{1 – \sqrt {\cos 2x} }}{{{x^2}}}}}.$ Mà: $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} – \sqrt[3]{{1 + 3x}}}}{{{x^2}}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} – (1 + x)}}{{{x^2}}}$ $ + \mathop {\lim }\limits_{x \to 0} \frac{{(x + 1) – \sqrt[3]{{1 + 3x}}}}{{{x^2}}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{ – 1}}{{\sqrt {1 + 2x} + x + 1}}$ $ + \mathop {\lim }\limits_{x \to 0} \frac{{x + 3}}{{{{(x + 1)}^2} + (x + 1)\sqrt[3]{{1 + 3x}} + \sqrt[3]{{{{\left( {1 + 3x} \right)}^2}}}}}$ $ = – \frac{1}{2} + 1 = \frac{1}{2}.$ $\mathop {\lim }\limits_{x \to 0} \frac{{1 – \sqrt {\cos 2x} }}{{{x^2}}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos 2x}}{{{x^2}}}$$.\frac{1}{{1 + \sqrt {\cos 2x} }}$ $ = 1.$ Vậy $B = \frac{1}{2}.$ Ví dụ 14. Tìm các giới hạn sau: 1. $A = \mathop {\lim }\limits_{x \to 0} {x^3}\sin \frac{1}{{{x^2}}}.$ 2. $B = $ $\mathop {\lim }\limits_{x \to + \infty } \left( {2\sin x + {{\cos }^3}x} \right)\left( {\sqrt {x + 1} – \sqrt x } \right).$ 1. Ta có: $0 \le \left| {{x^3}\sin \frac{1}{{{x^2}}}} \right| \le {x^3}.$ Mà $\mathop {\lim }\limits_{x \to 0} {x^3} = 0$ $ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left| {{x^3}\sin \frac{1}{{{x^2}}}} \right| = 0$ $ \Rightarrow \mathop {\lim }\limits_{x \to 0} {x^3}\sin \frac{1}{{{x^2}}} = 0.$ Vậy $A = 0.$ 2. Ta có: $B = \mathop {\lim }\limits_{x \to + \infty } \frac{{2\sin x + {{\cos }^3}x}}{{\sqrt {x + 1} + \sqrt x }}.$ Mà $0 \le \left| {\frac{{2\sin x + {{\cos }^2}x}}{{\sqrt {x + 1} + \sqrt x }}} \right|$ $ \le \frac{3}{{\sqrt {x + 1} + \sqrt x }} \to 0$ khi $x \to + \infty .$ Do đó: $B = 0.$
