习题2.3

第 1 题

证明本节的性质1与性质2.

解

第 2 题

证明定理2.3.1.

解

第 3 题

设 \(\lim\limits_{x \to x_0}f(x) = A\), 证明:

(1) \(\lim\limits_{x \to x_0}f^2(x) = A^2\);

(2) \(\lim\limits_{x \to x_0}\sqrt{f(x)} = \sqrt{A}(A > 0)\);

(3) \(\lim\limits_{x \to x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}\)

解

第 4 题

若 \(\lim\limits_{x \to x_0}f(x) = A > B\), 则 \(\exists \delta > 0\), 当 \(x \in U(x_0, \delta)\) 时, \(f(x) > B\).

解

第 5 题

证明定理2.3.2.

解

第 6 题

求下列极限(其中各题中的 \(m\) 与 \(n\) 均为正整数).

(1) \(\lim\limits_{x \to 2}(5-3x)(3x-1)\);

(2) \(\lim\limits_{x \to \frac{\pi}{2}}\frac{\sin x}{x}\);

(3) \(\lim\limits_{x \to 2}\frac{x^2 - 2x}{x^2 - 3x + 2}\);

解

(1) \(\lim\limits_{x \to 2}(5-3x)(3x-1) = (5 - 3 \times 2)(3\times 2 - 1) = -5\)

(2) \(\lim\limits_{x \to \frac{\pi}{2}}\frac{\sin x}{x} = \frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}} = \frac{2}{\pi}\)

(3) \(\lim\limits_{x \to 2}\frac{x^2 - 2x}{x^2 - 3x + 2} = \lim\limits_{x \to 2}\frac{x(x - 2)}{(x - 1)(x - 2)} = \lim\limits_{x \to 2}\frac{x}{x - 1} = \frac{2}{2 - 1} = 2\)