6.2 Answer

6.2

1.

(a) $D = \{(x,y) : 0 < x^2 + y^2 < 1\}$ represents a unit circle and does not contain the center and the perimeter. Thus, take any point in $D$, the neighborhood can be included in $D$．Thus，$D$ is open set．

$D$ is included in the circle with the radius $R>1$. Thus, bounded．

Any two points in $D$ are connected by a continuous curve inside of $D$. Thus, connected.

A connected open set is called a region. Thus $D$ is region.

The boundary of $D$ is $${\partial D = \{(x,y) : x^2 + y^2 = 1\}}$$

The closure of $D$ is $${ {\bar D} = \{(x,y) : 0 \leq x^2 + y^2 \leq 1 \}}$$

(b)

Since $D = \{(x,y) : xy \leq 0\}$, we have $\sim D = \{(x,y) : xy > 0\}$．Now $\sim D$ is open set. Thus $D$ is closed set．

For any open disk $\{(x,y) : x^2 + y^2 < R\}$, $D$ is not included. Thus unbounded．

Any two point in $D$ can be connected by continuous curve inside of $D$. Thus $D$ is connected.

The boundary of $D$ is $${\partial D = \{(x,y) : xy = 0\}}$$

The closure of $D$ is $${{\bar D} = \{(x,y) : 0 \leq xy \leq 0 \}}$$

2.

(a) The least degree in the numerator $= 1 <$ the least degree in the denominator $=2$．Thus, as $x$ goes to 0, the denominator goes to 0 faster than the numerator．So, we let $y = mx$. Then

$$\lim_{x \to 0}\frac{\sqrt{mx^2}}{x^2 + m^2 x^2} = \lim_{x \to 0}\frac{\sqrt{m}}{x(1+m^2)} = \infty$$

Therefore, no limit exist．

(b) The least degree in the numerator $=$ the least degree in the denominator $=2$. Thus we let $y = mx$. Then

$$\lim_{x \to 0}\frac{m x^2}{x^2 + m^2 x^2 + m^4 x^4} = \lim_{x \to 0}\frac{\sqrt{m}}{(1+m^2)} = \frac{m}{1 + m^2}.$$

Note that this limit depends on the values of $m$. Thus, $\lim_{(x,y) \to 0}\frac{xy}{x^2 + y^2 + y^4}$ does not exist. (c) The least degree in the numerator $= 2 >$ the least degree in the denominator $=1$. Thus, we let $x = r\cos{\theta}, y = r\sin{\theta}$. Then

$$\vert\frac{xy}{x^2 + y^2 + y}\vert = \vert\frac{r^2 \cos{\theta}\... ... + r\sin{\theta}}\vert = \vert\frac{1}{1 + \frac{1}{r}\sin{\theta}}\vert \to 0.$$

Therefore,by the squeezing theorem, $\lim_{(x,y) \to 0}\frac{xy}{x^2 + y^2 + y} = 0$．

3.

(a) Check to see

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

．

The least degree in the numerator $= 3 >$ the least degree in the denominator $=2$．Thus the numerator goes to 0 faster than the denominator．Now let $x = r\cos{\theta}, y = r\sin{\theta}$. Then

$$\vert\frac{x^2 y}{x^2 + y^2}\vert = \vert\frac{r^3 \cos^{2}{\thet... ...= \vert r\cos^{2}{\theta}\sin{\theta}\vert \leq \vert r\vert \to 0 \ (r \to 0).$$

Thus， $\lim_{(x,y) \to 0}\frac{x^2 y}{x^2 + y^2} = 0 = f(0,0)$ and $f(x,y)$ is continuous at $(0,0)$．

(b) The least degree in the numerator $=$ the least degree in the denominator $=2$. Then we let $y = mx$.

$$\lim_{x \to 0}\frac{x^2 - m^2 x^2}{x^2 + m^2 x^2} = \lim_{x \to 0}\frac{1 - m^2}{(1 + m^2)} = \frac{1 - m^2}{1 + m^2}.$$

This limit depends on the values of $m$. Thus, $\lim_{(x,y) \to 0}\frac{x^2 - y^2}{x^2 + y^2}$ does not exist．Therefore, $f(x,y)$ is dicontinuous at $(0,0)$.

(c) When $(x,y)$ approaches $(0,0)$，the speed of approaching 0 is faster than the $\log(x^2 + y^2)$ approaches $\infty$. Then let $x = r\cos{\theta}, y = r\sin{\theta}$.

$$0 \leq \vert f(x,y)\vert = \vert 2r^2 \cos{\theta}\sin{\theta}\log{r}\vert \leq \vert 2r^2 \log{r}\vert \to 0 \ (r \to 0).$$

Thus, $\lim_{(x,y) \to 0}f(x,y) = 0 \neq f(0,0) = -1$ and $f(x,y)$ is discontinuous at$(0,0)$．