다변수함수의 극한과 연속

(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

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