Limit of Functions of Two Variables

For a function $z = f(x,y)$, if $(x,y) \rightarrow (x_{0},y_{0})$, then $f(x,y) \rightarrow l$. Then we call $l$ a limit of a function $f(x,y)$ as $(x,y)$ approaches $(x_{0},y_0)$ and denote

$$\lim_{(x,y) \rightarrow (x_{0},y_0)}f(x,y) = l.$$

For every $\varepsilon > 0$, there exists $\delta > 0$ such that whenever $0 < \sqrt{(x - x_{0})^2 + (y - y_{0})^2} < \delta$, then $\vert f(x,y) - l\vert < \varepsilon$. Then

$$\lim_{(x,y) \rightarrow (x_{0},y_0)}f(x,y) = l$$

NOTE Note that $\sqrt{(x - x_{0})^2 + (y - y_{0})^2} < \delta$ means that there are points $(x,y)$ in the circle of the center $(x_0, y_0)$ and the radius less that $\delta$. The relation $\vert f(x,y) - l\vert < \varepsilon$ implies that the value of $f(x,y)$ and $l$ is very close to each other.

Example 4..4   Find the following limit.

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

SOLUTION As $(x,y) \rightarrow (0,0)$, $x^2 + y^2 \rightarrow 0, x^{2}y \rightarrow 0$. Now we must find out which one gets close to 0 faster. To do this, we compare the least degree in $x$ and $y$ of the numerator and the denominator. Note that the least degree in $x$ and $y$ in the numerator is 3 and the least degree in $x$ and $y$ in the denominator is 2. Then we have a better chance of convergence. Thus we show that every approach to $(0,0)$, we have a unique number to converge. Let

$$x = r\cos{\theta}, y = r\sin{\theta}$$

Then

$$0 \leq \vert\frac{x^{2}y}{x^2 + y^2}\vert = \vert\frac{r^3 \cos^2{\theta}\sin{\theta}}{r^2}\vert \leq \vert r\vert$$

Note that $\lim_{(x,y) \to (0,0)}0 = 0$ and $\lim_{r \to 0}\vert r\vert = 0$. Thus by squeezing theorem, we have

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

and

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

Exercise 4..4   Find the following limit.

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

SOLUTION Let $y = mx^k$. Then we have $${\frac{mx^{k+1}}{x^2 + m^2 x^{2k}}}$$. Thus we let $k = 1$ and $y = mx$.

$$\frac{xy}{x^2 + y^2} = \frac{mx^2}{x^2(1 + m^2)} = \frac{m}{1 + m^2} (x \neq 0)$$

Thus

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

Now note that this value depends on the value of $m$. If $m=0$, then the limit is 0 and if $m=1$, then the limit is $\frac{1}{2}$. Thus as $(x,y) \rightarrow (0,0)$, the limit does not exist $\blacksquare$

$$\lim_{(x,y) \rightarrow (x_0,y_0)}f(x,y) = f(x_0,y_0)$$

then we say $f((x,y))$ is continuous at $(x,y) = (x_0,y_0)$. If $f(x,y)$ is continuous at all points of $D \subset {\mathbb{R}}^2$, then we say $f(x,y)$ is continuous on $D$.

NOTE Note that $f(x,y)$ is continuous at $(x_0, y_0)$ if and only if
1. $\lim_{(x,y) \rightarrow (x_0,y_0)}f(x,y)$ exists
2. $f(x_0,y_0)$ is defined
3. $\lim_{(x,y) \rightarrow (x_0,y_0)}f(x,y) = f(x_0,y_0)$

Theorem 4..1   [If

$$\lim_{(x,y) \rightarrow (x_0,y_0)}f(x,y) = f(x_0,y_0), \lim_{(x,y) \rightarrow (x_0,y_0)}g(x,y) = g(x_0,y_0),$$

then

1. $${\lim_{(x,y) \rightarrow (x_0,y_0)}\big(f(x,y) \pm g(x,y)\big) = f(x_0,y_0) \pm g(x_0, y_0)}$$
2. $${\lim_{(x,y) \rightarrow (x_0,y_0)}k f(x,y) = k f(x_0,y_0)\hskip 0.5cm {\rm provide} k {\rm is constant}}$$
3. $${\lim_{(x,y) \rightarrow (x_0,y_0)}f(x,y)g(x,y) = f(x_0,y_0)g(x_0,y_0)}$$
4. $${\lim_{(x,y) \rightarrow (x_0,y_0)}\frac{f(x,y)}{g(x,y)} = \frac{f(x_0,y_0)}{g(x_0,y_0)}\hskip 0.5cm (g(x_0,y_0) \neq 0)}$$

Example 4..5   Determine the following function is continuous at $(0,0)$.

$$f(x,y) = \left\{\begin{array}{cl} \frac{x y}{\sqrt{x^2 + y^2}}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.$$

SOLUTION Note that $f(x,y)$ is continuous except for the denominator is 0. Thus we only need to check continuity at $(0,0)$. The least degree of the numerator is 2 and the least degree of the denominator is 1. Thus we let $x = r\cos{\theta}, y = r\sin{\theta}$. Then

$$0 \leq \vert\frac{xy}{\sqrt{x^2+y^2}}\vert = \vert\frac{r^2\cos{\theta}\sin{\theta}}{r}\vert \leq \vert r\cos{\theta}\sin{\theta}\vert \leq \vert r\vert \longrightarrow 0 (r \rightarrow 0)$$

Therefore, $\lim_{(x,y) \rightarrow (0,0)}f(x,y) = 0$. Since $f(0,0) = 0$, $f(x,y)$ is continuous at $(0,0)$ $\blacksquare$

Exercise 4..5   Determine the following function is continuous at $(0,0)$.

$$f(x,y) = \left\{\begin{array}{cl} \frac{x^2y}{x^4+y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array} \right.$$

SOLUTION Note that $f(x,y)$ is continuous except for the denominator is 0. Thus we only need to check continuity at $(0,0)$. The least degree of the numerator is 3 and the least degree of $x$ in the denominator is 4. Then we let $y = mx^2$ and

$$\frac{x^2y}{x^4+y^2} = \frac{mx^4}{(1+m^2)x^4} = \frac{m}{1+m^2} (x \neq 0)$$

Thus $\lim_{(x,y) \rightarrow (0,0)}\frac{x^2y}{x^4+y^2} = \frac{n}{1+m^2} = \left\{\begin{array}{ll} 0 & m = 0\\ \frac{1}{2} & m = 1\end{array}\right.$. Thus, $f(x,y)$ is not continuous at $(0,0)$ $\blacksquare$

Exercise A

1.

Find the limit of the following functions．

(a) $${\lim_{(x,y) \to (1,1)}\frac{x - y + 1}{x + y - 1}}$$ (b) $${\lim_{(x,y) \to (0,0)}\frac{2x - 3y}{x + y}}$$ (c) $${\lim_{(x,y) \to (0,0)}\frac{x^{2} - y^{2}}{x + y}}$$

2.

Determine the following functions are continuous at $(0,0)$．

(a) $${f(x,y) = \frac{xy}{x^{2} + y^{2} + 1}}$$ (b) $${ f(x,y) = \left\{\begin{array}{cl} \frac{x^2}{x^2+y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.}$$

(c) $${ f(x,y) = \left\{\begin{array}{cl} \frac{x^2y}{x^4+y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.}$$

Exercise B

2.

Find the limit of the following functions as $(x,y) \rightarrow (0,0)$．

(a) $${\frac{\sqrt{xy}}{x^2 + y^2}}$$ (b) $${\frac{xy}{x^2 + y^2 + y^4} }$$ (c) $${\frac{xy}{x^2 + y^2 + y}}$$

3.

Determine whether the following functions are continuous at $(0,0)$ or not．

(a) $${f(x,y) = \left\{\begin{array}{cl} \frac{x^2y}{x^2+y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.}$$

(b) $${ f(x,y) = \left\{\begin{array}{cl} \frac{x^2-y^2}{x^2+y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.}$$

(c) $${ f(x,y) = \left\{\begin{array}{cl} xy \log(x^2 + y^2), & (x,y) \neq (0,0)\\ -1, & (x,y) = (0,0) \end{array}\right.}$$