다변수함수의 극한과 연속

The graph increases without bound as $x\,\text{and}\,y$ both approach zero.

        ![On xyz space, there is a shape drawn that decreases to 0 as x and y increase or decrease but that increases greatly closer to the origin. It increases to such an extent that the graph is cut off above z = 10, which coincides with a circle of radius 0.6 around (0, 0, 10).](openstax/CNX_Calc_Figure_14_02_201.jpg)

(a) In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, so that

$$\begin{aligned} \frac{x^8+y^8}{x^4+y^4} &=\frac{r^8\cos^8\theta+r^8\sin^8\theta}{r^4\cos^4\theta+r^4\sin^4\theta} =r^4\frac{\cos^8\theta+\sin^8\theta}{\cos^4\theta+\sin^4\theta} \end{aligned}$$

As

$$\begin{aligned} \frac{\cos^8\theta+\sin^8\theta}{\cos^4\theta+\sin^4\theta} &\le \frac{\cos^8\theta+2\cos^4\theta\sin^4\theta+\sin^8\theta} {\cos^4\theta+\sin^4\theta} =\frac{\big(\cos^4\theta+\sin^4\theta\big)^2}{\cos^4\theta+\sin^4\theta} \\ &=\cos^4\theta+\sin^4\theta \le 2 \end{aligned}$$

we have

$$0\le \frac{x^8+y^8}{x^4+y^4}\le 2r^4$$

As $\displaystyle \lim_{(x,y)\to (0,0)}2r^4=0$, the squeeze theorem yields $\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^8+y^8}{x^4+y^4}=0$.

(b) In polar coordinates

$$\begin{aligned} \frac{xy^5}{x^8+y^{10}} &=\frac{r^6\cos\theta\,\sin^5\theta}{r^8\cos^8\theta+r^{10}\sin^{10}\theta} =\frac{1}{r^2}\frac{\cos\theta\,\sin^5\theta}{\cos^8\theta+r^2\sin^{10}\theta} \end{aligned}$$

As $(x,y)\to (0,0)$ the first fraction $\frac{1}{r^2}\to\infty$ but the second factor can take many different values. For example, if we send $(x,y)$ towards the origin along the $y$–axis, i.e. with $\theta=\pm\frac{\pi}{2}$,

$$\begin{aligned} \lim_{\genfrac{}{}{0pt}{}{(x,y)\to(0,0)}{x=0}}\frac{xy^5}{x^8+y^{10}} =\lim_{y\to 0} \frac{0}{y^{10}}=0 \end{aligned}$$

but if we send $(x,y)$ towards the origin along the line $y=x$, i.e. with $\theta=\frac{\pi}{4},\frac{5\pi}{4}$,

$$\begin{aligned} \lim_{\genfrac{}{}{0pt}{}{(x,y)\to(0,0)}{y=x} }\frac{xy^5}{x^8+y^{10}} =\lim_{x\to 0} \frac{x^6}{x^8+x^{10}} =\lim_{x\to 0} \frac{1}{x^2}\frac{1}{1+x^2} =+\infty \end{aligned}$$

and if we send $(x,y)$ towards the origin along the line $y=-x$, i.e. with $\theta=-\frac{\pi}{4},\frac{3\pi}{4}$,

$$\begin{aligned} \lim_{\genfrac{}{}{0pt}{}{(x,y)\to(0,0)}{y=-x} }\frac{xy^5}{x^8+y^{10}} =\lim_{x\to 0} \frac{-x^6}{x^8+x^{10}} =\lim_{x\to 0}- \frac{1}{x^2}\frac{1}{1+x^2} =-\infty \end{aligned}$$

So $\frac{xy^5}{x^8+y^{10}}$ does not approach a single value as $(x,y)\to(0,0)$ and the limit does not exist.

(a) In polar coordinates

$$\frac{x^3-y^3}{x^2+y^2}=\frac{r^3\cos^3\theta-r^3\sin^3\theta}{r^2} =r\cos^3\theta-r\sin^3\theta$$

Since

$$\big|r\cos^3\theta-r\sin^3\theta\big|\le 2r$$

and $2r\rightarrow 0$ as $r\rightarrow 0$, the limit exists and is $0$.

(b) The limit as we approach $(0,0)$ along the $x$-axis is

$$\begin{aligned} \lim_{t\rightarrow 0}\frac{x^2-y^4}{x^2+y^4}\bigg|_{(x,y)=(t,0)} =\lim_{t\rightarrow 0}\frac{t^2-0^4}{t^2+0^4} =1 \end{aligned}$$

On the other hand the limit as we approach $(0,0)$ along the $y$-axis is

$$\begin{aligned} \lim_{t\rightarrow 0}\frac{x^2-y^4}{x^2+y^4}\bigg|_{(x,y)=(0,t)} =\lim_{t\rightarrow 0}\frac{0^2-t^4}{0^2+t^4} =-1 \end{aligned}$$

These are different, so the limit as $(x,y)\rightarrow 0$ does not exist.

We can gain a more detailed understanding of the behaviour of $\frac{x^2-y^4}{x^2+y^4}$ near the origin by switching to polar coordinates.

$$\frac{x^2-y^4}{x^2+y^4} =\frac{r^2\cos^2\theta-r^4\sin^4\theta}{r^2\cos^2\theta+r^4\sin^4\theta} =\frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta}$$

Now fix any $\theta$ and let $r\rightarrow 0$ (so that we are approaching the origin along the ray that makes an angle $\theta$ with the positive $x$-axis). If $\cos\theta\ne 0$ (i.e. the ray is not part of the $y$-axis)

$$\begin{aligned} \lim_{r\rightarrow 0} \frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta} =\frac{\cos^2\theta}{\cos^2\theta} =1 \end{aligned}$$

But if $\cos\theta= 0$ (i.e. the ray is part of the $y$-axis)

$$\begin{aligned} \lim_{r\rightarrow 0} \frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta} =\lim_{r\rightarrow 0} \frac{-r^2\sin^4\theta}{r^2\sin^4\theta} =\frac{-\sin^4\theta}{\sin^4\theta} =-1 \end{aligned}$$

(a) In polar coordinates $x=r\cos\theta$, $y=r\sin\theta$

$$\begin{aligned} \frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2} &=\frac{2r^2\cos^2\theta + r^3\cos^2\theta\sin\theta - r^3\cos\theta\sin^2\theta + 2r^2\sin^2\theta}{r^2} \\ &=2+ r\big[\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big] \end{aligned}$$

As

$$r\big|\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big| \le 2r \rightarrow 0\text{ as }r\rightarrow 0$$

we have

$$\lim_{(x,y)\rightarrow(0,0)} \frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}=2$$

(b) Since

$$\begin{aligned} \frac{x^2y^2 -2 x^2y + x^2} {(x^2 + y^2-2y+1)^2} =\frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} \end{aligned}$$

and, in polar coordinates centred on $(0,1)$, $x=r\cos\theta$, $y=1+r\sin\theta$,

$$\frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} =\frac{r^4\cos^2\theta\sin^2\theta}{r^4} =\cos^2\theta\sin^2\theta$$

we have that the limit does not exist. For example, if we send $(x,y)$ to $(0,1)$ along the line $y=1$, so that $\theta=0$, we get the limit $0$, while if we send $(x,y)$ to $(0,1)$ along the line $y=x+1$, so that $\theta=\frac{\pi}{4}$, we get the limit $\frac{1}{4}$.