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Multivariable Epsilon-Delta proof example

Multivariable epsilon-delta proofs are generally harder than their single variable counterpart. The difficulty comes from the fact that we need to manipulate $|f(x,y) - L|$ into something of the form $\sqrt{(x-a)^2 + (y-b)^2}$, which is much harder to do than the simple $|x-a|$ with single variable proofs. This often requires several ingenious algebraic tricks that, at first don't seem to make much sense. Consider the following example.

Let \(f : \{(x,y)\in {\mathbb{R}}^2 : x^2 \ne y^2\} \to \mathbb{R}\) be defined by

\begin{equation*} f(x,y) = \frac{x^3 - y^3}{x^2-y^2}. \end{equation*}

Prove that the limit

\begin{equation*} \lim_{(x,y)\to (a,a)} f(x,y) \end{equation*}

exists for all real $a \ne 0$.

First, observe that we may factor a $x+y$ term,

\begin{equation*} f(x,y) = \frac{(x-y)(x^2+xy + y^2)}{(x-y)(x+y)} = \frac{x^2+xy + y^2}{x+y} \end{equation*}

so we determine that the limit approaching $(a,a)$ should probably be this expression evaluated at $(a,a)$. This gives us $L = 3a/2$. We prove this using the epsilon-delta definition.

Let \({\epsilon}> 0\) be given and choose \(\delta = \min\{{\epsilon}/3, |a|\}\) with \(a\ne 0\). Let \((x,y) \in \text{dom}(f) = \{(x,y)\in {\mathbb{R}}^2 : x^2 \ne y^2\}\) such that \(0 < |(x,y) - (a,a)| < \delta\). Then

\begin{equation*} \begin{aligned} {\left | f(x,y) - 3a/2 \right |} &= {\left | \frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)} -3a/2 \right |} \\[3.5ex] &= {\left | \frac{(x+y)^2 - xy}{x+y} - 3a/2 \right |} \\[3.5ex] &= {\left | (x-a) + (y-a) + a/2 - \frac{xy}{x+y} \right |} \\[3.5ex] &= {\left | (x-a) + (y-a) + \frac{(x+y)a/2 - xy}{x+y} \right |} \\[3.5ex] &= {\left | (x-a) + (y-a) + \frac{-x(y-a)/2 - y(x-a)/2}{x+y} \right |} \\[3.5ex] &\leq |x-a| + |y-a| + \frac{1}{2}{\left | \frac{x(y-a) + y(x-a)}{x+y} \right |} \\[3.5ex] &\leq |x-a| + |y-a| + \frac{1}{2}{\left | \frac{x(y-a)}{x+y} \right |} + \frac{1}{2}{\left | \frac{y(x-a)}{x+y} \right |} \\[3.5ex] &\leq |x-a| + |y-a| + \frac{1}{2}|y-a| + \frac{1}{2}|x-a| \quad\quad \text{(*)} \\[3.5ex] &= \frac{3}{2}|x-a| + \frac{3}{2}|y-a| \\[3.5ex] &\leq 3\sqrt{(x-a)^2 +(y-a)^2} < 3\delta = {\epsilon}. \end{aligned} \end{equation*}

To obtain $(\text{*})$, we used the fact that because \(|(x,y) - (a,a)| < \delta \leq |a|\), therefore \(x\ne 0\) and \(y\ne 0\) and \(\text{sgn}(x) = \text{sgn}(y)\). This in turn implies that

\begin{equation*} \frac{1}{|x+y|} < \frac{1}{|x|} \quad \text{and} \quad \frac{1}{|x+y|} < \frac{1}{|y|}. \end{equation*}

This completes the proof.

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