다변수함수의 극한과 연속

(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 $\frac{1}{4}$.

(a) Since, in polar coordinates,

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

we have that the limit does not exist. For example,

• if we send $(x,y)$ to $(0,0)$ along the positive $x$-axis, so that $\theta=0$, we get the limit $\sin\theta\cos\theta\big|_{\theta=0}=0$,
• while if we send $(x,y)$ to $(0,0)$ along the line $y=x$ in the first quadrant, so that $\theta=\frac{\pi}{4}$, we get the limit $\sin\theta\cos\theta\big|_{\theta=\pi/4}=\frac{1}{2}$.

(b) This limit does not exist, since if it were to exist the limit

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

would also exist. (Recall that $\displaystyle \lim_{t\rightarrow 0}\tfrac{\sin t}{t} =1$.)

(c) Since

$$\begin{aligned} \lim_{(x,y)\rightarrow(-1,1)}\big[x^2+2xy^2+y^4\big] &=(-1)^2+2(-1)(1)^2+(1)^4=0 \\ \lim_{(x,y)\rightarrow(-1,1)}\big[1+y^4\big] &=1+(1)^4=2 \end{aligned}$$

and the second limit is nonzero,

$$\lim_{(x,y)\rightarrow(-1,1)}\frac{x^2+2xy^2+y^4}{1+y^4}=\frac{0}{2}=0$$

(d) Since the limit along the positive $x$-axis

$$\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}|y|^x\Big|_{(x,y)=(t,0)} =\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}0^t =\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}0 =0$$

and the limit along the $y$-axis

$$\lim_{t\rightarrow 0}|y|^x\Big|_{(x,y)=(0,t)} =\lim_{t\rightarrow 0}|t|^0 =\lim_{t\rightarrow 0}1 =1$$

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