05. 다변수함수의 극한과 연속Limits and Continuity of Multivariable Functions

Solution.

(1) $f(x,y)=\sqrt{y-x^2}$ is defined exactly when $y-x^2\ge0$, i.e., $y\ge x^2$, so the domain is $D=\{(x,y)\in\mathbb R^2 : y\ge x^2\}$. Fixing $x=0$, $y-x^2=y$ takes every value in $[0,\infty)$, so $f=\sqrt{y-x^2}$ also takes every value in $[0,\infty)$. Hence the range is $[0,\infty)$.

(2) Since $g(x,y)=y-x^2$ is continuous, $D^c=\{(x,y):y\lt x^2\}=g^{-1}((-\infty,0))$ is the preimage of an open interval, hence open, so $D$ is closed. On the other hand, for a boundary point $(a,a^2)\in D$ on the parabola, every neighborhood contains points with $y\lt x^2$, so no neighborhood of $(a,a^2)$ lies entirely in $D$. Hence $D$ is not open.

(3) $D$ contains every point $(0,y)$ with $y\ge0$, so points of $D$ can be arbitrarily far from the origin. Thus no open ball of finite radius contains $D$, so $D$ is not bounded.

(4) Since $z=\sqrt{y-x^2}\ (z\ge0)$, squaring gives $z^2=y-x^2$, i.e., $y=x^2+z^2$. This is the half ($z\ge0$) of the paraboloid of revolution $y=x^2+z^2$ about the $y$-axis. At each height $y=c\ (c\ge0)$ the cross-section is the semicircle $x^2+z^2=c,\ z\ge0$ of radius $\sqrt c$, and the surface tapers to the single point at the origin when $y=0$ — a half-paraboloid trough shape.

(5) Let $f(x,y)=c\ (c\ge0)$. Then $\sqrt{y-x^2}=c \Leftrightarrow y-x^2=c^2 \Leftrightarrow y=x^2+c^2$. So the level curves form a family of parabolas $y=x^2+c^2$ with vertex $(0,c^2)$, nested along the $y$-axis; at $c=0$ this is exactly the domain boundary $y=x^2$.

Solution.

(1) $z=1/xy$ is defined exactly when $xy\ne0$, i.e., $x\ne0$ and $y\ne0$, so the domain is the plane minus both coordinate axes, $\{(x,y)\in\mathbb R^2 : x\ne0,\ y\ne0\}$. Since $xy$ can be any nonzero real number, the range is $\mathbb R\setminus\{0\}$.

(2) $xy$ is defined for every pair of reals, so the domain is all of $\mathbb R^2$. Fixing $y=1$, $xy=x$ takes every real value, so $\sin xy$ takes every value in $[-1,1]$. Hence the range is $[-1,1]$.

(3) Since $x^2+y^2+z^2\ge0$ always holds, the domain is all of $\mathbb R^3$. At $(t,0,0)$, $w=\sqrt{t^2}=|t|$, so as $t$ ranges over $[0,\infty)$, $w$ takes every value in $[0,\infty)$. Hence the range is $[0,\infty)$.

(4) $x^2+y^2+z^2=0$ holds only at $(x,y,z)=(0,0,0)$, so the domain is $\mathbb R^3\setminus\{(0,0,0)\}$. On the domain, $x^2+y^2+z^2$ takes every value in $(0,\infty)$, and $t\mapsto1/t$ is a bijection on $(0,\infty)$, so the range of $w$ is $(0,\infty)$.

Solution.

(1) Let $A=\{(x,y):x^2+y^2\le1\}$. The interior is $\mathrm{int}\,A=\{(x,y):x^2+y^2\lt1\}$ (the open unit disk), and the boundary is $\partial A=\{(x,y):x^2+y^2=1\}$ (the unit circle). Since $A$ contains all its boundary points, $A\ne\mathrm{int}\,A$, so $A$ is not open. On the other hand, the complement $A^c=\{x^2+y^2\gt1\}$ is the preimage of an open interval under the continuous function $x^2+y^2$, hence open, so $A$ is closed.

(2) The interior of $\mathbb R^2$ is $\mathbb R^2$ itself (every point is interior), and its boundary is $\varnothing$. $\mathbb R^2$ is open as the whole space, and its complement $\varnothing$ is also open, so $\mathbb R^2$ is closed as well. Hence $\mathbb R^2$ is both open and closed.

(3) Let $B=\{(x,y):x,y\in\mathbb Q\}$. Since $\mathbb Q$ is dense in $\mathbb R$, every neighborhood of any point of $B$ contains points with an irrational coordinate, so no point of $B$ is interior; the interior is $\varnothing$. Also $B$ is dense in $\mathbb R^2$, so $\overline B=\mathbb R^2$, and $B^c$ (points with at least one irrational coordinate) is dense as well, so $\overline{B^c}=\mathbb R^2$; hence the boundary is $\partial B=\overline B\cap\overline{B^c}=\mathbb R^2$. Since $B\ne\mathrm{int}\,B=\varnothing$, $B$ is not open, and since $B^c$ is also dense (no point of $B^c$ is interior to $B^c$, so $B^c$ is not open), $B$ is not closed either.

(4) The interior and boundary of $\varnothing$ are both $\varnothing$. $\varnothing$ is (vacuously) open, and since its complement $\mathbb R^2$ is open, $\varnothing$ is also closed. Hence $\varnothing$ is both open and closed.

Solution. $f(x,y)=\ln(9-x^2-y^2)$ is defined exactly when $9-x^2-y^2\gt0$, i.e., $x^2+y^2\lt9$, so the domain is the open disk of radius $3$ centered at the origin, $\{(x,y):x^2+y^2\lt9\}$. Every point of this set is at distance less than $3$ from the origin, so the domain is contained entirely inside the (closed) ball of radius $3$. Hence the domain is bounded.

Note. The original problem prints $f(x,y)=\sqrt{x^2+y^2+z^2}$, but since this is a level-surface problem and $z$ appears under the square root, this is evidently a misprint for $f(x,y,z)=\sqrt{x^2+y^2+z^2}$. We solve the corrected problem $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ below.

Solution. Let $f=c\ (c\ge0)$. Then $\sqrt{x^2+y^2+z^2}=c \Leftrightarrow x^2+y^2+z^2=c^2$. For $c\gt0$ this is the sphere of radius $c$ centered at the origin, and for $c=0$ it degenerates to the single point at the origin. Hence the level surfaces of $f$ form a family of concentric spheres centered at the origin, with radius increasing as $c$ increases.

Solution. The level curves of $f_1(x,y)=xy$ are $xy=c$. For $c\ne0$ these are hyperbolas not passing through the origin, lying in quadrants I and III when $c\gt0$ and in quadrants II and IV when $c\lt0$, while $c=0$ gives the two coordinate axes themselves. This matches figure B (a hyperbolic saddle pattern in each quadrant).

The level curves of $f_2(x,y)=\cos(x-y)$ are $\cos(x-y)=c$, i.e., $x-y=\pm\cos^{-1}c+2k\pi\ (k\in\mathbb Z)$, a family of parallel lines of slope $1$. This matches figure C (parallel diagonal stripes).

The level curves of $f_3(x,y)=x^2+y^2$ are $x^2+y^2=c\ (c\ge0)$, circles of radius $\sqrt c$ centered at the origin. This matches figure A (concentric circles).

Hence the matching is (1) $z=xy \to$ figure B, (2) $z=\cos(x-y) \to$ figure C, (3) $z=x^2+y^2 \to$ figure A.

Solution.

(1) As $(x,y)\to(0,1)$, the denominator $x^2y+5xy-y^3$ equals $0+0-1=-1\ne0$ at $(0,1)$, so this rational function is continuous at $(0,1)$ and we may substitute directly. The numerator equals $0-0+3=3$ at $(0,1)$, so the limit is $\dfrac{3}{-1}=-3$.

(2) $\sqrt{x^2+y^2}$ is continuous on all of $\mathbb R^2$, so substituting directly gives $\sqrt{3^2+(-4)^2}=\sqrt{25}=5$.

(3) On the region $x,y\ge0,\ x\ne y$ where $\sqrt x,\sqrt y$ are defined, $$\frac{x^2-xy}{\sqrt x-\sqrt y}=\frac{x(x-y)}{\sqrt x-\sqrt y}=\frac{x(x-y)(\sqrt x+\sqrt y)}{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}=\frac{x(x-y)(\sqrt x+\sqrt y)}{x-y}=x(\sqrt x+\sqrt y).$$ The right-hand side is continuous at $(0,0)$, so the limit is $0\cdot(\sqrt0+\sqrt0)=0$.

(4) Since $\dfrac{y^2}{x^2+y^2}\le1$, we have $\left|\dfrac{4xy^2}{x^2+y^2}\right|=4|x|\cdot\dfrac{y^2}{x^2+y^2}\le4|x|$, and $4|x|\to0$ as $(x,y)\to(0,0)$, so by the Squeeze Theorem the limit is $0$.

(5) Along the path $y=kx$ ($k$ constant), $\dfrac yx=\dfrac{kx}{x}=k$ depends on the path. For instance, along $y=0$ the limit is $0$, while along $y=x$ the limit is $1$ — different values. Hence the limit diverges (does not exist).

(6) Along the path $y=kx^2$ ($k$ constant), $$\frac{2x^2y}{x^4+y^2}=\frac{2x^2\cdot kx^2}{x^4+k^2x^4}=\frac{2kx^4}{(1+k^2)x^4}=\frac{2k}{1+k^2},$$ independent of $x$ but depending on $k$. For instance, $k=0$ ($y=0$) gives limit $0$, while $k=1$ ($y=x^2$) gives limit $1$ — different values. Hence the limit diverges.

Solution. Along the path $y=kx$ ($k$ constant) approaching the origin, $$f(x,kx)=\frac{2x\cdot kx}{x^2+k^2x^2}=\frac{2k}{1+k^2},$$ which depends on $k$ (e.g., $0$ along $y=0$, and $1$ along $y=x$). Hence $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ takes different values along different paths, so it does not exist. Since the limit fails to exist, $f$ is not continuous at the origin, regardless of the value $f(0,0)=0$.

Solution.

(1) Let $u=2x-y$. As $(x,y)\to(2,0)$ with $2x-y\ne4$, we have $u\to4,\ u\ne4$, and $$\frac{\sqrt{2x-y}-2}{2x-y-4}=\frac{\sqrt u-2}{u-4}=\frac{\sqrt u-2}{(\sqrt u-2)(\sqrt u+2)}=\frac1{\sqrt u+2}.$$ As $u\to4$, $\dfrac1{\sqrt u+2}\to\dfrac1{2+2}=\dfrac14$, so the limit is $\dfrac14$.

(2) At $(1,3,4)$, where $x,y,z$ are all nonzero, $\dfrac1x+\dfrac1y+\dfrac1z$ is continuous, so substituting directly gives $1+\dfrac13+\dfrac14=\dfrac{12+4+3}{12}=\dfrac{19}{12}$.

(3) Along the path $y=1$ as $x\to1$, $h_1(x)=\tan1-\tan x$ satisfies $h_1(1)=0$, so $$\lim_{x\to1}\frac{\tan1-\tan x}{1-x}=\lim_{x\to1}\frac{\tan x-\tan1}{x-1}=\left.\frac{d}{dx}\tan x\right|_{x=1}=\sec^21.$$ Along the path $x=1$ as $y\to1$, $h_2(y)=\tan y-y\tan1$ satisfies $h_2(1)=\tan1-\tan1=0$, so $$\lim_{y\to1}\frac{\tan y-y\tan1}{y-1}=h_2'(1)=\sec^21-\tan1.$$ Since $\tan1\ne0$, we have $\sec^21\ne\sec^21-\tan1$, so the two paths give different limits and the original limit diverges.

(4) Since $|\sin(1/x)|\le1$, we have $|y\sin(1/x)|\le|y|$, and $|y|\to0$ as $(x,y)\to(0,0)$, so by the Squeeze Theorem the limit is $0$.

(5) Let $t=xy$. Since $|xy|\le\dfrac{x^2+y^2}{2}$, we have $t\to0$ as $(x,y)\to(0,0)$. The function $g(t)=\dfrac{\tan^{-1}t}{t}$, defined at $t=0$ by $g(0):=1$, is continuous there (from the Taylor expansion $\tan^{-1}t=t-\frac{t^3}{3}+\cdots$, $g(t)\to1$), so the limit is $1$.

(6) Since $x^2\le x^2+y^2+z^2$, we have $0\le\dfrac{x^2z^2}{x^2+y^2+z^2}\le z^2$, and $z^2\to0$ as $(x,y,z)\to(0,0,0)$, so by the Squeeze Theorem $\dfrac{x^2z^2}{x^2+y^2+z^2}\to0$. Since $\cos$ is continuous, the limit is $\cos0=1$.

Solution. By the definition, $f(x,y)=0$ only at points with $0\lt y\lt x^4$, and $f(x,y)=1$ everywhere else.

(1) Near $(0,1)$, $x$ is close to $0$ so $x^4$ is close to $0$, while $y$ is close to $1$; for instance, throughout the neighborhood $|x|\lt0.8,\ y\gt\frac12$ we have $x^4\lt0.8^4\approx0.41\lt\frac12\lt y$, so $y\ge x^4$ holds. Hence $f\equiv1$ on this neighborhood, so the limit is $1$.

(2) At $(2,3)$, $x^4=16$, so $y=3\lt16=x^4$ and $y=3\gt0$. Thus $(2,3)$ is an interior point of the region $0\lt y\lt x^4$ (the "otherwise" case), and on a neighborhood of $(2,3)$ where $x^4$ stays near $16$ and $y$ stays near $3$, the inequality $0\lt y\lt x^4$ still holds. Hence $f\equiv0$ on this neighborhood, so the limit is $0$.

(3) Along the path $y=0$ ($x\to0,\ x\ne0$), $y\le0$ always holds, so $f(x,0)=1$ and the limit along this path is $1$. Along the path $y=x^4/2$ ($x\to0,\ x\ne0$), $0\lt\frac{x^4}2\lt x^4$, so $f\left(x,\frac{x^4}2\right)=0$ and the limit along this path is $0$. Since the two paths give different limits, $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ diverges.

Solution. For $(x,y)\ne(0,0)$, since $|xy|\le\dfrac{x^2+y^2}2$ and $|x^2-y^2|\le x^2+y^2$, $$\left|xy\,\frac{x^2-y^2}{x^2+y^2}\right|\le\frac{\frac{x^2+y^2}2\cdot(x^2+y^2)}{x^2+y^2}=\frac{x^2+y^2}2.$$ Since the right-hand side tends to $0$ as $(x,y)\to(0,0)$, the Squeeze Theorem gives $$\lim_{(x,y)\to(0,0)}xy\,\frac{x^2-y^2}{x^2+y^2}=0.$$ Hence $f$ is continuous at the origin exactly when $f(0,0)=0$.