习题1.3

第 1 题

下列函数当 \((x, y) \to (0, 0)\) 时, 其极限是否存在? 若存在, 求出极限.

(1) \(\dfrac{\arcsin(x^2 + y^2)}{x^2 + y^2}\);

(3) \((x^2 + y^2)e^{-x-y}\);

(5) \(\dfrac{x^2 - y^2}{x^2 + y^2}\);

(7) \(\dfrac{x^3 - y^3}{x + y}\);

(9) \(\dfrac{x^2}{x^2 + y^2}\);

(11) \(\dfrac{x^4y^4}{(x^2 + y^4)^3}\);

解

(1) 当 \((x, y)\to(0, 0)\) 时 \(x^2 + y^2 \to 0\). 因此 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{\arcsin(x^2 + y^2)}{x^2 + y^2} = \lim\limits_{u\to 0}\dfrac{\arcsin u}{u} = 1\).

(3) \(\lim\limits_{(x, y)\to(0, 0)}(x^2 + y^2)e^{-x-y} = \lim\limits_{(x, y)\to(0, 0)}(x^2 + y^2)\lim\limits_{(x, y)\to(0, 0)}e^{-x-y} = 0\).

(5) 当 \((x, y)\) 沿直线 \(x = 0\) 趋于 \((0, 0)\) 时, \(f(x, y) = \frac{-y^2}{y^2} = -1\). 当 \((x, y)\) 沿直线 \(y = 0\) 趋于 \((0, 0)\) 时, \(f(x, y) = \frac{x^2}{x^2} = 1\). 两者不相等, 故极限不存在.

(7) 当 \((x, y)\) 沿直线 \(y = kx^3 - x\) 趋于 \((0, 0)\) 时, \(f(x, y) = \dfrac{x^3 - (kx^3 - x)^3}{x + (kx^3 - x)} = \dfrac{-k^3x^9 + 3k^2x^7 - 3kx^5 + 2x^3}{kx^3} = \dfrac{-k^3x^6 + 3k^2x^4 - 3kx^2 + 2}{k} \to \dfrac{2}{k} \quad (x \to 0)\). 因此极限随着 \(k\) 的变化而变化, 故极限不存在.

(9) 当 \((x, y)\) 沿直线 \(x = 0\) 趋于 \((0, 0)\) 时, \(f(x, y) = \dfrac{1}{y^2} \to +\infty, \quad y \to 0\). 故极限不存在.

(11) 当 \((x, y)\) 沿直线 \(y = x\) 趋于 \((0, 0)\) 时, \(f(x, y) = \dfrac{x^8}{(x^2 + x^4)^3} > \dfrac{1}{x^4} \to +\infty, \quad x \to 0\). 故极限不存在.

第 2 题

求下列函数极限.

(1) \(\lim\limits_{x\to 3 \atop y\to 0}\dfrac{\ln(x + \sin y)}{\sqrt{x^2 + y^2}}\);

(3) \(\lim\limits_{x\to +\infty \atop y\to -\infty}(x^2 + y^2)e^{y - x}\);

(5) \(\lim\limits_{x \to \infty \atop y \to \infty}\left(\dfrac{\left|xy\right|}{x^2 + y^2}\right)^{x^2}\)

解

(1) \(\lim\limits_{x\to 3 \atop y\to 0}\dfrac{\ln(x + \sin y)}{\sqrt{x^2 + y^2}} = \dfrac{\ln(x + \sin y)}{\sqrt{x^2 + y^2}}\Bigg\vert_{x = 3\atop y = 0} = \dfrac{\ln 3}{3}\).

(3) \(\lim\limits_{x\to +\infty \atop y\to -\infty}(x^2 + y^2)e^{y - x} = \lim\limits_{x\to +\infty \atop y\to -\infty}e^y\cdot\dfrac{x^2}{e^x} + \lim\limits_{x\to +\infty \atop y\to -\infty} \dfrac{1}{e^x}\cdot y^2e^y = 0 \cdot 0 + 0 \cdot 0 = 0\)

(5) 由于 \(0 \le \left(\dfrac{\left|xy\right|}{x^2 + y^2}\right)^{x^2} \le \left(\dfrac{1}{2}\right)^{x^2}\) 且 \(\lim\limits_{x \to \infty}\left(\dfrac{1}{2}\right)^{x^2} = 0\), 故由夹逼准则知 \(\lim\limits_{x \to \infty \atop y \to \infty}\left(\dfrac{\left|xy\right|}{x^2 + y^2}\right)^{x^2} = 0\).

第 3 题

讨论下列累次极限与二重极限是否存在, 若存在求值.

(1) \(\lim\limits_{x\to \infty}\lim\limits_{y\to \infty}\sin \dfrac{\pi x}{2x + y}, \lim\limits_{y\to \infty}\lim\limits_{x\to \infty}\sin \dfrac{\pi x}{2x + y}, \lim\limits_{x\to \infty \atop y \to \infty}\sin \dfrac{\pi x}{2x + y};\)

(2) \(\lim\limits_{x\to +\infty}\lim\limits_{y\to 0^{+}}\dfrac{x^y}{1+x^y}, \lim\limits_{y\to 0^{+}}\lim\limits_{x\to +\infty}\dfrac{x^y}{1+x^y}, \lim\limits_{x\to +\infty \atop y \to 0^{+}}\dfrac{x^y}{1+x^y}\);

(3) \(\lim\limits_{x\to 0}\lim\limits_{y\to 0}(x+y)\sin \dfrac{1}{x}\sin \dfrac{1}{y}, \lim\limits_{y\to 0}\lim\limits_{x\to 0}(x+y)\sin \dfrac{1}{x}\sin \dfrac{1}{y}, \lim\limits_{x\to 0 \atop y \to 0}(x+y)\sin \dfrac{1}{x}\sin \dfrac{1}{y}\).

解

(2) \(\lim\limits_{x\to +\infty}\lim\limits_{y\to 0^{+}}\dfrac{x^y}{1+x^y} = \lim\limits_{x\to +\infty}\dfrac{1}{1+1} = \dfrac{1}{2}\). \(\lim\limits_{y\to 0^{+}}\lim\limits_{x\to +\infty}\dfrac{x^y}{1+x^y} = \lim\limits_{y\to 0^{+}}1 = 1\). 由 1.3.1 节结论知 \(\lim\limits_{x\to +\infty \atop y \to 0^{+}}\dfrac{x^y}{1+x^y}\) 不存在.

(3) 显然累次极限均不存在. 因为 \(0 \le \left|(x+y)\sin\dfrac{1}{x}\sin\dfrac{1}{y}\right| \le \left|x + y\right|\), 且 \(\lim\limits_{x \to 0 \atop y \to 0}\left|x+y\right| \to 0\), 故由夹逼准则知 \(\lim\limits_{x\to 0 \atop y \to 0}(x+y)\sin \dfrac{1}{x}\sin \dfrac{1}{y} = 0\).

第 4 题

(1) 举例说明累次极限存在性与二重极限的存在性互不包含;

(2) 证明: 若二元函数 \(f\) 在某一点的两个累次极限和二重极限都存在, 则这三个值相等.

解

(2) 由于 \(f(x, y)\) 的二重极限存在, 不妨设 \(\lim\limits_{(x, y)\to(x_0, y_0)}f(x, y) = A\). 再设 \(g(x) = \lim\limits_{y\to y_0}f(x, y)\), 即证 \(\lim\limits_{x \to x_0}g(x) = A\). 由定义知 \(\forall \epsilon > 0, \exists \delta > 0\) 使得在 \(0 < \sqrt{(x - x_0)^2 + (y - y_0)^2} < \delta\) 时有 \(\left|f(x, y) - A\right| < \dfrac{\epsilon}{2}\). 左右两侧取极限可知 \(\lim\limits_{y\to y_0}\left|f(x, y)- A\right| \le \dfrac{\epsilon}{2}\). 由于绝对值函数是连续函数, 因此可以转化为 \(\left|\lim\limits_{y \to y_0}f(x, y) - A\right| \le \dfrac{\epsilon}{2} < \epsilon\), 即 \(\left|g(x) - A\right| <\epsilon\). 观察此时 \(\delta\) 的取值范围, 发现有 \(0 < \left|x - x_0\right| \le \sqrt{(x - x_0)^2 + (y - y_0)^2} < \delta\). 故由定义可知 \(\lim\limits_{x \to x_0}g(x) = A\). 此即 \(\lim\limits_{x\to x_0}\lim\limits_{y\to y_0}f(x, y) = \lim\limits_{(x, y)\to(x_0, y_0)}f(x, y)\). 由对称性可知 \(\lim\limits_{y\to y_0}\lim\limits_{x\to x_0}f(x, y) = \lim\limits_{(x, y)\to(x_0, y_0)}f(x, y)\). 命题得证.

第 5 题

用定义证明函数 \(f(x, y) = \sqrt{x^2 + y^2}\) 在 \(\mathbb{R}^2\) 上连续.

解

第 6 题

判断下列函数的在 \((0, 0)\) 点的连续性.

(1) \(f(x, y) = \begin{cases}\dfrac{\sin(x^3 + y^3)}{x^2+y^2}, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0;\end{cases}\)

(2) \(f(x, y) = \begin{cases}1, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0;\end{cases}\)

(3) \(f(x, y) = \begin{cases}\dfrac{x^2y^2}{(x^2+y^2)^{\frac{3}{2}}}, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0;\end{cases}\)

(4) \(f(x, y) = \begin{cases}\dfrac{xy^2}{x^2+y^4}, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0.\end{cases}\)

解

(1) 因为 \(0 \le \left|\dfrac{\sin(x^3 + y^3)}{x^2 + y^2}\right| \le \left|\dfrac{x^3 + y^3}{x^2 + y^2}\right| = \left|(x + y)\left(1 + \dfrac{xy}{x^2 + y^2}\right)\right| \le \dfrac{3}{2}(x + y)\), 且 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{3}{2}(x + y) = 0\), 故由夹逼准则知 \(\lim\limits_{(x, y)\to(0, 0)}f(x, y) = 0 = f(0, 0)\). 因此该函数在 \((0, 0)\) 处连续.

(2) 因为 \(\lim\limits_{(x, y)\to(0, 0)}f(x, y) = 1 \neq 0 = f(0, 0)\), 所以该函数在 \((0, 0)\) 处不连续.

(3) 因为 \(0 \le \left|\dfrac{x^2y^2}{(x^2+y^2)^{\frac{3}{2}}}\right| \le \left|\dfrac{x^2 + y^2}{(2xy)^{\frac{3}{2}}}\right| = \left|\dfrac{\sqrt{xy}}{2\sqrt{2}}\right|\), 且 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{\sqrt{xy}}{2\sqrt{2}} = 0\), 故由夹逼准则知 \(\lim\limits_{(x, y)\to(0, 0)}f(x, y) = 0 = f(0, 0)\). 因此该函数在 \((0, 0)\) 处连续.

(4) 考虑极限 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{xy^2}{x^2+y^4}\). 当 \((x, y)\) 沿着 \(y = k\sqrt{x}\) 趋于 \((0, 0)\) 时, 有 \(f(x, y) = f(x, k\sqrt{x}) = \dfrac{k^2}{1 + k^4}\) 随着 \(k\) 的变化而变化. 因此该极限不存在. 故该函数在 \((0, 0)\) 处不连续.

第 7 题

考察下列函数在平面上的连续性, 并指出在哪些点上函数是连续的.

(1) \(f(x, y) = \begin{cases}\dfrac{x - y^2}{x^3+y^3}, &x + y \neq 0, \\ 0, &x + y = 0;\end{cases}\)

(2) \(f(x, y) = \begin{cases}\dfrac{x}{y^2}e^{-\frac{x^2}{y^2}}, &y \neq 0, \\ 0, &y = 0;\end{cases}\)

解

(1) 当 \(x + y \neq 0\) 的时候显然函数是连续的. 现只需考察 \(x + y = 0\) 时的情况即可. 当 \(x + y = 0\) 且 \((x, y)\) 不趋向于 \((0, 0)\) 时, 显然分子不为 \(0\), 但分母趋向于 \(0\), 极限不存在, 因此在 \(x + y = 0(x \neq 0)\) 的区域内不连续. 最后再考察 \((x, y)\to (0, 0)\) 的情况. 假设 \((x, y)\) 沿着 \(y = kx\) 趋于 \((0, 0)\), 则 \(\lim\limits_{(x, y)\to(0, 0)}f(x, y) = \lim\limits_{x \to 0}\dfrac{x - (kx)^2}{x^3 + (kx)^3} = \lim\limits_{x \to 0}\dfrac{1 - k^2x}{(1 + k^3)x^2} = \lim\limits_{x \to 0}\dfrac{-k^2}{2x(1 + k^3)}\) 极限不存在. 因此函数在 \((0, 0)\) 处也不连续. 综上所述, 函数在除了 \(x + y = 0\) 这条直线外的区域都连续.

(2) 先固定 \(x = x_0\), \(f(x, y) = f(x_0, y) = \dfrac{x_0}{y^2e^{\frac{x_0^2}{y^2}}} \to 0, \quad y \to 0\). 再考虑当 \((x, y)\to(0, 0)\) 时的极限值. 不妨设 \((x, y)\) 沿着 \(y = kx\) 趋于 \((0, 0)\), 则 \(\lim\limits_{(x, y)\to (0, 0)}f(x, y) = \lim\limits_{x\to 0}\dfrac{x}{(kx)^2}e^\frac{1}{k^2}\) 不存在. 因此该函数在 \((0, 0)\) 处不连续, 在平面上其他区域内都连续.

第 8 题

设 \(f\) 是定义在 \(\mathbb{R}^2\) 上的连续函数, 且 \(\lim\limits_{x^2 + y^2 \to \infty}f(x, y) = +\infty\), 证明: \(f\) 有最小值.

证明

记 \(f(0, 0) = A\). 因为 \(\lim\limits_{x^2 + y^2 \to \infty}f(x, y) = +\infty\), 所以 \(\exists r_0 > 0\) 使得 \(\forall (x, y)\) 满足 \(x^2 + y^2 > r_0^2\), 都有 \(f(x, y) > A\). 记 \(D = \{(x, y) | x^2 + y^2 \le r_0^2\}\) 为一有界闭集, 故 \(D\) 中可以取到最小值. 不妨设 \(f(x_0, y_0)\) 为 \(D\) 中最小值. 由于 \((0, 0) \in D\), 则 \(f(x_0, y_0) \le f(0, 0) = A\). 所以 \(\forall (x, y) \in \mathbb{R}^2, f(x_0, y_0) \le f(x, y)\). 所以 \(f\) 在 \(\mathbb{R}^2\) 上存在最小值.

第 9 题

当 \((x, y) \to (0, 0)\) 时, \(f(x, y) = o(\rho^m), g(x, y) = o(\rho^n)\), 其中 \(\rho = \sqrt{x^2 + y^2}\), 证明:

(1) \(f(x, y) + g(x, y) = o(\rho^k), k = \min\{m, n\}\),

(2) \(f(x, y)g(x, y) = o(\rho^{m+n})\).

证明

(1) 由题知 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{f(x, y)}{\rho^m} = 0, \lim\limits_{(x, y)\to(0, 0)}\dfrac{g(x, y)}{\rho^n} = 0\). 不妨设 \(m \le n\). 因此 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{f(x, y) + g(x, y)}{\rho^m} = \lim\limits_{(x, y)\to (0, 0)}\left(\dfrac{f(x, y)}{\rho^m} + \rho^{n - m}\dfrac{g(x, y)}{\rho^n}\right) = 0 + 0 = 0\). 由定义知 \(f(x, y) + g(x, y) = o(\rho^k)\), 其中 \(k = \min\{m, n\}\).

(2) 由题知 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{f(x, y)}{\rho^m} = 0, \lim\limits_{(x, y)\to(0, 0)}\dfrac{g(x, y)}{\rho^n} = 0\). 因此 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{f(x, y)g(x, y)}{\rho^{m + n}} = \lim\limits_{(x, y)\to (0, 0)}\left(\dfrac{f(x, y)}{\rho^m} \cdot \dfrac{g(x, y)}{\rho^n}\right) = 0 \cdot 0 = 0\). 由定义知 \(f(x, y)g(x, y) = o(\rho^{m+n})\).

第 10 题

当 \((x, y)\to(0, 0)\) 时, 讨论下列无穷小的阶(若有阶, 求阶; 若无阶, 说明理由).

(1) \(\sin (x^2 + y^2)\);

(2) \(\ln(1 + \sqrt{x^2 + y^2})\);

(3) \((x^2 + y^2)\sin\dfrac{1}{\sqrt{x^2 + y^2}}\);

(4) \(x + y + 2xy\);

解

(1) 记 \(\sqrt{x^2 + y^2} = \rho\). 则 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{\sin (x^2 + y^2)}{x^2 + y^2} = \lim\limits_{\rho \to 0}\dfrac{\sin\rho^2}{\rho^2} = 1\). 故为 \(2\) 阶无穷小.

(2) 记 \(\sqrt{x^2 + y^2} = \rho\). 则 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{\ln(1 + \sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} = \lim\limits_{\rho \to 0}\dfrac{\ln(1 + \rho)}{\rho} = 1\). 故为 \(1\) 阶无穷小.

(3) 记 \(\sqrt{x^2 + y^2} = \rho\). 则 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{(x^2 + y^2)\sin\dfrac{1}{\sqrt{x^2 + y^2}}}{(\sqrt{x^2 + y^2})^k} = \lim\limits_{\rho \to 0}\dfrac{\rho^2\sin\dfrac{1}{\rho}}{\rho^k}\) 不可能为一固定常数, 所以无阶.

(4) 由于 \(\lim\limits_{(x, y)\to(0, 0)}\dfrac{x + y + 2xy}{(\sqrt{x^2 + y^2})^k}\) 不可能为一固定常数, 所以无阶.