# Proving Limits of Functions of Two Variables

Recall that for a two variable real-valued function $z = f(x, y)$, then $\lim_{(x, y) \to (a,b)} f(x, y) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid f(x, y) - L \mid < \epsilon$. We will now use the definition to prove that some value $L$ is the limit as $(x, y) \to (a, b)$.

## Example 1

**Let $f(x, y) = k$. Prove that $\lim_{(x, y) \to (a,b)} k = k$.**

First note that the function $f(x, y) = k$ represents the plane $z = k$ which is parallel and elevated $k$ units above/below the $xy$-plane. Thus, every point on this plane has $z$-coordinate $k$. Thus, if $(x, y) \to (a,b)$, we can intuitively see that $z = f(x, y) \to k$.

Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \epsilon$ then $\mid f(x, y) - k \mid < \epsilon$.

We start with $\mid f(x,y) - k \mid = \mid k - k \mid = 0$. But $\epsilon > 0$, and so for any chosen $\delta$ we have that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \epsilon$ then $\mid f(x, y) - k \mid < \epsilon$.