# 2024年考研数二第05题解析：二元函数在一点处可微的判定、有界震荡无极限

## 一、题目

(A) $\frac{\partial f(x, y)}{\partial x}$ 连续, $f(x, y)$ 可微

(B) $\frac{\partial f(x, y)}{\partial x}$ 连续, $f(x, y)$ 不可微

(C) $\frac{\partial f(x, y)}{\partial x}$ 不连续, $f(x, y)$ 可微

(D) $\frac{\partial f(x, y)}{\partial x}$ 不连续, $f(x, y)$ 不可微

## 二、解析

$$f_{x}^{\prime}(0,0)=\lim \limits_{x \rightarrow 0} \frac{f(x, 0)-f(0,0)}{x}=\lim \limits_{x \rightarrow 0} \frac{0-0}{x}=0$$

$$f_{y}^{\prime}(0,0) = \lim \limits_{y \rightarrow 0} \frac{f(0, y)-f(0,0)}{y} = \lim \limits_{y \rightarrow 0} \frac{0 – 0}{y} = 0$$

\begin{aligned} f_{x}^{\prime}(x, y) \\ \\ & = \left[ \left(x^{2}+y^{2}\right) \sin \frac{1}{x y} \right]^{\prime}_{x} \\ \\ & = 2x \sin \frac{1}{xy} + (x^{2} + y^{2}) \frac{1}{y} \cdot \frac{-1}{x^2} \cos \frac{1}{xy} \\ \\ & = 2 x \sin \frac{1}{x y} – \frac{x^{2} + y^{2}}{x^{2} y} \cos \frac{1}{x y} \end{aligned}

$$\lim \limits_{(x, y) \rightarrow(0,0)} \frac{f(x, y)-f(0,0)-f_{x}^{\prime}(0,0) x-f_{y}^{\prime}(0,0) y}{\sqrt{x^{2}+y^{2}}}=$$

$$\lim \limits_{(x, y) \rightarrow(0,0)} \frac{f(x, y) – 0 – 0 \cdot x – 0 \cdot y}{\sqrt{x^{2}+y^{2}}}=$$

$$\lim \limits_{(x, y) \rightarrow(0,0)} \frac{\left(x^{2}+y^{2}\right) \sin \frac{1}{x y}}{\sqrt{x^{2}+y^{2}}}=$$

$$\lim \limits_{(x, y) \rightarrow(0,0)} \sqrt{x^{2}+y^{2}} \sin \frac{1}{x y}=$$

$$0 \cdot \text{有界震荡函数} = 0$$