Proving Limits of Functions of Two Variables

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 $\mid f(x, y) - k \mid < \epsilon$, and so $\lim_{(x, y) \to (a,b)} k = k$.

Example 2

Prove that $\lim_{(x, y) \to (0, 0)} 2x + 3y = 0$.

Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then $\mid 2x + 3y \mid < \epsilon$.

Note that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then we have that both $\mid x \mid < \delta$ and $\mid y \mid < \delta$. Let $\delta = \frac{\epsilon}{5}$. Then we have that:

(1)
\begin{align} \quad \mid 2x + 3y \mid ≤ 2 \mid x \mid + 3 \mid y \mid < 2 \delta + 3 \delta = 5 \delta = 5 \cdot \frac{\epsilon}{5} = \epsilon \end{align}

Therefore $\lim_{(x, y) \to (0, 0)} 2x + 3y = 0$.

Example 3

Prove that $\lim_{(x, y) \to (0, 0)} 2x^3 - 3xy + 1 = 1$.

Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then $\mid 2x^3 - 3xy + 1 - 1 \mid = \mid 2x^3 - 3xy \mid < \epsilon$.

Once again, note that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then we have that both $\mid x \mid < \delta$ and $\mid y \mid < \delta$. Let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{5} \}$. Then since $\delta ≤ 1$ we have that $\delta^n ≤ 1$ for $n = 1, 2, 3, ...$ and so:

(2)
\begin{align} \quad \mid 2x^3 - 3xy \mid ≤ 2 \mid x^3 \mid + 3 \mid xy \mid = 2 \mid x \mid^3 + 3 \mid x \mid \mid y \mid < 2 \delta^3 + 3 \delta^2 < 2 \delta + 3 \delta = 5 \delta < \epsilon \end{align}

Therefore $\lim_{(x, y) \to (0, 0)} 2x^3 - 3xy + 1 = 1$.