Section 14.2: Limits and Continuity

Definition
Let \(f\) be a function of two variables whose domain \(D\) includes points arbitrarily close to \((a, b)\). Then we say that the limit of \(f(x, y)\) as \((x, y)\) approaches \((a, b)\) is \(L\) and we write:

\[ \lim_{(x, y) \to (a, b)} f(x, y) = L \]

if for every number \(\varepsilon > 0\) there is a corresponding number \(\delta > 0\) such that

\[ \text{if } (x, y) \in D \quad \text{and} \quad 0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \quad \text{then} \quad |f(x, y) - L| < \varepsilon \]

Simple Interpretation Function \(f(x,y)\) value approaches \(L\) regardless of the path through which \((x,y)\) approaces \((a,b)\).

Limit Does not Exist if Find two paths along which function values approaches different limit.

Example
Show that \(\lim_{(x, y) \to (0, 0)} \frac{x^2 - y^2}{x^2 + y^2}\) does not exist.

Example
Show that \(\lim_{(x, y) \to (0, 0)} \frac{xy}{x^2 + y^2}\) does not exist.

Example
Show that \(\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^4}\) does not exist.

Example
Show that \(\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2}\) exists.

Note Detailed proof is not in the syllabus. Just plot the graph. See that it is pretty smooth at \((0,0)\).

Definition
A function \(f\) of two variables is called continuous at \((a, b)\) if

\[ \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) \]

We say \(f\) is continuous on \(D\) if \(f\) is continuous at every point \((a, b)\) in \(D\).

Note In this course you will not be asked to prove continuity. Use these rules.

• Polynomials are always continuous
• Most forumulas as continuous if you can evaluate the formula at the given point.

Example
Evaluate
\[ \lim_{(x, y) \to (1, 2)} \left(x^2 y^3 - x^3 y^2 + 3x + 2y\right). \]

Solution
Since \(f(x, y) = x^2 y^3 - x^3 y^2 + 3x + 2y\) is a polynomial, it is continuous everywhere, so we can find the limit by direct substitution:

\[ \lim_{(x, y) \to (1, 2)} \left(x^2 y^3 - x^3 y^2 + 3x + 2y\right) = 1^2 \cdot 2^3 - 1^3 \cdot 2^2 + 3 \cdot 1 + 2 \cdot 2 = 11 \]

Example
Where is the function
\[ f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \]
continuous?

Solution
The function \(f\) is discontinuous at \((0, 0)\) because it is not defined there. Since \(f\) is a rational function, it is continuous on its domain, which is the set
\[ D = \{(x, y) \mid (x, y) \neq (0, 0)\}. \]