다변수 미분 전체 문제

(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) 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)

$$\begin{aligned} f_x(x,y,z)&=3x^2y^4z^5 & f_x(0,-1,-1)&=0\\ f_y(x,y,z)&=4x^3y^3z^5 & f_y(0,-1,-1)&=0\\ f_z(x,y,z)&=5x^3y^4z^4 & f_z(0,-1,-1)&=0 \end{aligned}$$

(b)

$$\begin{aligned} w_x(x,y,z)&=\frac{yz e^{xyz}}{1+e^{xyz}} & w_x(2,0,-1)&=0\\ w_y(x,y,z)&=\frac{xz e^{xyz}}{1+e^{xyz}} & w_y(2,0,-1)&=-1\\ w_z(x,y,z)&=\frac{xy e^{xyz}}{1+e^{xyz}} & w_z(2,0,-1)&=0 \end{aligned}$$

(c)

$$\begin{aligned} f_x(x,y)&=-\frac{x}{(x^2+y^2)^{3/2}} & f_x(-3,4)&=\frac{3}{125}\\ f_y(x,y)&=-\frac{y}{(x^2+y^2)^{3/2}} & f_y(-3,4)&=-\frac{4}{125} \end{aligned}$$