# 2015年考研数二第03题解析

## 题目

$$A. \alpha – \beta > 1$$

$$B. 0 < \alpha – \beta \leqslant 1$$

$$C. \alpha – \beta > 2$$

$$D. 0 < \alpha – \beta \leqslant 2$$

## 解析

$$\lim_{x \rightarrow 0^{+}} x^{2} \cos \frac{1}{x} =$$

$$\lim_{x \rightarrow 0^{+}} \frac{\cos \frac{1}{x}}{x^{-2}} \Rightarrow$$

$\cos \frac{1}{x}$ 是一个有界函数，$x^{-2}$ 趋向于无穷大，一个有界函数除以一个无穷大则等于零，于是有：

$$\lim_{x \rightarrow 0^{+}} \frac{\cos \frac{1}{x}}{x^{-2}} = 0.$$

$$\lim_{x \rightarrow 0^{+}} x^{3} \cos \frac{1}{x} =$$

$$\lim_{x \rightarrow 0^{+}} \frac{\cos \frac{1}{x}}{x^{-3}} = 0.$$

$$\lim_{x \rightarrow 0^{+}} x^{4} \cos \frac{1}{x} =$$

$$\lim_{x \rightarrow 0^{+}} \frac{\cos \frac{1}{x}}{x^{-4}} = 0.$$

$$f^{‘}(0) =$$

$$\lim_{x \rightarrow 0} \frac{f(x) – f(0)}{x – 0} =$$

$$\lim_{x \rightarrow 0} \frac{f(x)}{x} =$$

$$\lim_{x \rightarrow 0} x^{\alpha – 1} \cos \frac{1}{x^{\beta}}.$$

$$0 \cdot 有界函数 = 0.$$

$$\infty \cdot 有界函数 = \infty$$

$$\alpha – 1 \geqslant 0 \Rightarrow$$

$$\alpha \geqslant 1.$$

$$f^{‘}(x) =$$

$$\alpha x^{\alpha – 1} \cos \frac{1}{x^{\beta}} + x^{\alpha} (- \sin \frac{1}{x^{\beta}}) (- \beta x^{- \beta – 1}) =$$

$$\alpha x^{\alpha – 1} \cos \frac{1}{x^{\beta}} + x^{\alpha} (\sin \frac{1}{x^{\beta}}) (\beta x^{- \beta – 1}) =$$

$$\alpha x^{\alpha – 1} \cos \frac{1}{x^{\beta}} + (\sin \frac{1}{x^{\beta}}) (\beta x^{\alpha – \beta – 1}) ①$$

$$\alpha – 1 \geqslant 0;$$

$$\alpha – \beta – 1 \geqslant 0.$$

$$\alpha \geqslant 1;$$

$$\alpha – \beta \geqslant 1.$$

$$\alpha > 0, \beta > 0.$$

$$\alpha \geqslant 1.$$

$$\alpha – \beta \geqslant 1.$$

EOF