다변수함수의 극한과 연속

(a) Along $y=mx$, the limit is:

$$\begin{aligned} \lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} & = \lim_{x\to 0} \frac{\sin(x^2)}{mx} \\ \text{apply L'H\^opital's Rule} \\ &= \lim_{x\to 0} \frac{2x\cos(x^2)}{m}\\ &= 0. \end{aligned}$$ (b) Along $x=0$, the limit is: $$\lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} = \lim_{x\to 0} \frac{\sin(x^2)}{x^2}.$$ This can be evaluated with L'Hopital's Rule or from known limits; it is 1.

Since the limits along the lines $y=mx$ are not the same as the limit along $y=x^2$, the overall limit does not exist.

(a) Along $y=2$, the limit is:

$$\begin{aligned} \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} & = \lim_{x\to 1} \frac{x-1}{x^2-1}\\ &= \lim_{x\to 1} \frac{1}{x+1}\\ &= 1/2. \end{aligned}$$ (b) Along $y=x+1$, the limit is:

$$\begin{aligned} \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} & = \lim_{x\to 1} \frac{2(x-1)}{x^2-1}\\ &= \lim_{x\to 1} \frac{2}{x+1}\\ &= 1. \end{aligned}$$

Since the limits along the lines $y=2$ and $y=x+1$ differ, the overall limit does not exist.

(a) Along $x=\pi$, the limit is:

$$\begin{aligned} \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin x}{\cos y} & = \lim_{y\to \pi/2} \frac{0}{\cos y}\\ &= 0. \end{aligned}$$ (b) Along $y=x-\pi/2$, the limit is:

$$\begin{aligned} \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin x}{\cos y} & = \lim_{x\to \pi} \frac{\sin x}{\cos(x-\pi/2)}\\ \text{Apply L'H\^opital's Rule:} \\ &= \lim_{x\to \pi} \frac{\cos x}{-\sin(x-\pi/2)}\\ &= 1. \end{aligned}$$

Since the limits along the lines $x=\pi$ and $y=x-\pi$ differ, the overall limit does not exist.

(a) By inspection, when $y=0$, then $f(x,y)=1$ as long as $x \neq 0$. So, if we follow the $x$-axis in towards the origin, $f(x,y)=1$ along this route. (b) Also by inspection, when $x=0$, then $f(x,y)=0$ as long as $y \neq 0$. So, if we follow the $y$-axis in towards the origin, $f(x,y)=0$ along this route. (c) Since two different directions give us different values as we approach the origin, $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ does not exist.

(a) $\displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2$

(b) Switching to polar coordinates,

$$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r\cos\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{\cos\theta}{r} \end{aligned}$$

which does not exist, since, for example,

• if $\theta=0$, then

$$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{1}{r} =+\infty$$

• while if $\theta=\pi$, then

$$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{-1}{r} =-\infty$$

(c) Switching to polar coordinates,

$$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \cos^2\theta \end{aligned}$$

which does not exist, since, for example,

• if $\theta=0$, then

$$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 1 = 1$$

• while if $\theta=\frac{\pi}{2}$, then

$$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi/2}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 0 =0$$

(d) Switching to polar coordinates,

$$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^3\cos^3\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r\cos^3\theta =0 \end{aligned}$$

since $|\cos\theta|\le 1$ for all $\theta$.

(e) Switching to polar coordinates,

$$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta\ r^2\sin^2\theta} {r^2\cos^2\theta+r^4\sin^4\theta} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r^2\sin^2\theta \frac{\cos^2\theta}{\cos^2\theta+r^2\sin^4\theta} \\ &= 0 \end{aligned}$$

Here, we used that

$$\left|\sin^2\theta\frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\right| \le \frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta} \le \left.\begin{cases} \frac{\cos^2\theta} {\cos^2\theta}&\text{ if }\cos\theta\ne 0 \\ 0 &\text { if }\cos\theta =0 \end{cases} \right\} \le 1$$

for all $r>0$.

(f) To start, observe that

$$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] \end{aligned}$$

We may evaluate $\displaystyle \left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right]$ by l'Hopital's rule or by using the definition of the derivative to give

$$\lim_{x\rightarrow 0}\ \frac{\sin x}{x} =\lim_{x\rightarrow 0}\ \frac{\sin x-\sin 0}{x-0} =\frac{\mathrm{d}}{\mathrm{d}x}\sin x\bigg|_{x=0} =\cos x\Big|_{x=0}=1$$

Similarly, we may evaluate $\displaystyle \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right]$ by l'Hopital's rule or by using the definition of the derivative to give

$$\lim_{y\rightarrow 0}\ \frac{e^y-1}{y} =\lim_{y\rightarrow 0}\ \frac{e^y-e^0}{y-0} =\frac{\mathrm{d}}{\mathrm{d}y}e^y\bigg|_{y=0} =e^y\Big|_{y=0}=1$$

So all together

$$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] =[1]\ [1]=1 \end{aligned}$$

(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.