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

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

**Figure 4.7:** Understanding
$\includegraphics[width=3cm]{SOFTFIG-4/twovariablegraph_gr1.eps}$

Unlike the function of single variable, there are infinitely many approaches for $(x,y) \rightarrow (x_{0},y_0)$ . Thus in order to have the limit, for all $(x,y) \rightarrow (x_{0},y_0)$ , we must have $f(x,y) \rightarrow l$ .

**Figure 4.8:**
$\includegraphics[width=3.5cm]{SOFTFIG-4/approach.eps}$

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

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

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.

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

**Figure 4.9:** Example4-4
$\includegraphics[width=3.5cm]{SOFTFIG-4/degnum_gr1.eps}$

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.

Since $(x,y) \to (0,0)$ , larger the degree, faster to converge.

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

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

Then

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

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

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

and

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

Existence of Limit
To show $(x,y) \to (0,0)$ , it is enough to show $r \to 0$ .

Exercise 4..4 Find the following limit.

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

**Figure 4.10:** Exercise4-4
$\includegraphics[width=3.5cm]{SOFTFIG-4/denodeg_gr1.eps}$

Non existence
To show the limit does not exist, it is enough to show that two different approaches give a different limit. Nice way to do this is to set $y = mx^{k}$ and choose so that the least degree of the numerator and the least degree of the denominator are the same.

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

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

Thus

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

Now note that this value depends on the value of $m$

. If

, 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 $\blacksquare$

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

then we say $f((x,y))$

is continuous at $(x,y) = (x_0,y_0)$

. If

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

Understanding
A sum of continuous functions is continuous, A difference of continuous functions is continuous. A product of continuous functions is continuous, A quotient of continuous functions is continuous provided the denominator is not 0.

Theorem 4..1 [If

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

then

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

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

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

Enough to show $\lim_{(x,y) \rightarrow (0,0)}f(x,y) = f(0,0)$ .

**Figure 4.11:** Example4-5
$\includegraphics[width=3.5cm]{SOFTFIG-4/cont1_gr1.eps}$

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

$\displaystyle 0 \leq \vert\frac{xy}{\sqrt{x^2+y^2}}\vert$

$\displaystyle =$

$\displaystyle \vert\frac{r^2\cos{\theta}\sin{\theta}}{r}\vert \leq \vert r\cos{\theta}\sin{\theta}\vert \leq \vert r\vert \longrightarrow 0 (r \rightarrow 0)$

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

is continuous at $(0,0)$

$\blacksquare$

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

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

**Figure 4.12:** Exercise4-5
$\includegraphics[width=3.5cm]{SOFTFIG-4/cont1_gr2.eps}$

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

$\displaystyle \frac{x^2y}{x^4+y^2}$

$\displaystyle =$

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

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

2.

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

．

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

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

Exercise B

2.

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

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

3.

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

or not．

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

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

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