Limit Laws for Functions of Several Variables

# Limit Laws for Functions of Several Variables

We will now establish some of the extensions to the limit laws that we already know for single variable functions. All of the proofs for these theorems can be extended to $n$ variable real-valued functions too.

Theorem 1: Suppose that $\lim_{(x, y) \to (a,b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$. Then $\lim_{(x,y) \to (a,b)} [f(x,y) + g(x,y)] = L + M$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $\lim_{(x, y) \to (a,b)} f(x, y) = L$ then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ $\exists \delta_1 > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_1$ then $\mid f(x,y) - L \mid < \epsilon_1 = \frac{\epsilon}{2}$.

- Similarly, since $\lim_{(x, y) \to (a,b)} g(x,y) = M$ then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ $\exists \delta_2 > 0$ such that if $(x, y) \in D(g)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_2$ then $\mid g(x,y) - M \mid < \epsilon_2 = \frac{\epsilon}{2}$.

- We want to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x, y) \in D(f + g)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid [f(x, y) + g(x, y)] - [L + M] \mid < \epsilon$. Let $\delta = \mathrm{min} \{ \delta_1, \delta_2 \}$. Then we have that:

\begin{align} \quad \quad \mid [f(x, y) + g(x, y)] - [L + M] \mid ≤ \mid f(x,y) - L \mid + \mid g(x,y) - M \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $\lim_{(x,y) \to (a,b)} [f(x,y) + g(x,y)] = L + M$. $\blacksquare$

Theorem 2: Suppose that $\lim_{(x, y) \to (a,b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$. Then $\lim_{(x,y) \to (a,b)} [f(x,y) - g(x,y)] = L - M$. |

The proof of theorem 2 is very similar to that of theorem 1, so we will omit it.

Theorem 3: Suppose that $\lim_{(x, y) \to (a,b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$. Then $\lim_{(x,y) \to (a,b)} [f(x,y)][g(x,y)] = LM$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $\lim_{(x,y) \to (a,b)} f(x,y) = L$ then for $\epsilon_0 = 1 > 0$ $\exists \delta_0 > 0$ such that if $(x,y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_0$ then $\mid f(x,y) - L \mid < \epsilon_0 = 1$, and so this implies that for $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_0$ we have that $L - 1 < f(x,y) < L + 1$. Therefore for $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_0$ we have that $\mid f(x, y) \mid < 1 + \mid L \mid$.

- Now also, since $\lim_{(x,y) \to (a,b)} f(x,y) = L$ then for $\epsilon_1 = \frac{\epsilon}{2(1 + \mid M \mid)} > 0$ $\exists \delta_1 > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_1$ then we have that $\mid f(x,y) - L \mid < \epsilon_1 = \frac{\epsilon}{2(1 + \mid M \mid)}$.

- Similarly, since $\lim_{(x,y) \to (a,b)} g(x,y) = M$ then for $\epsilon_2 = \frac{\epsilon}{2(1 + \mid L \mid)} > 0$ $\exists \delta_2 > 0$ such that if $(x,y) \in D(g)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_2$ then we have that $\mid g(x,y) - M \mid < \epsilon_2 = \frac{\epsilon}{2(1 + \mid L \mid)}$.

- We want to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x,y) \in D(fg)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid f(x,y)g(x,y) - LM \mid < \epsilon$. Let $\delta = \mathrm{min} \{ \delta_0, \delta_1, \delta_2 \}$. Then we have that:

\begin{align} \quad \quad \mid f(x,y)g(x,y) - LM \mid = \biggr \rvert f(x,y)[g(x,y) - M] + M[f(x,y) - L] \biggr \rvert ≤ \mid f(x, y) \mid \mid g(x,y) - M \mid + \mid M \mid \mid f(x,y) - L \mid \\ < \quad \quad \mid f(x, y) \mid \mid g(x,y) - M \mid + (1 + \mid M \mid) \mid f(x,y) - L \mid < (1 + \mid L \mid) \frac{\epsilon}{2(1 + \mid L \mid)} + (1 + \mid M \mid) \frac{\epsilon}{\epsilon}{2(1 + \mid M \mid)} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $\lim_{(x,y) \to (a,b)} [f(x,y)][g(x,y)] = LM$. $\blacksquare$

Theorem 4: Suppose that $\lim_{(x, y) \to (a,b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$. Then $\lim_{(x,y) \to (a,b)} \frac{f(x,y)}{g(x,y)} = \frac{L}{M}$ provided that $M \neq 0$. |

The proof to theorem 4 can be obtained analogously to that of the quotient limit law proof for a single variable function.

Theorem 5: Suppose that $\lim_{(x, y) \to (a,b)} f(x, y) = L$ and if $k \in \mathbb{R}$ then $\lim_{(x, y) \to (a, b)} kf(x, y) = kL$. |

Once again, we will skip this proof as it is similar to the single variable case.

Theorem 6 (Squeeze Theorem): Suppose that $f(x,y) ≤ g(x,y) ≤ h(x,y)$ for $(x,y)$ near $(a,b)$. If $\lim_{(x,y) \to (a,b)} f(x,y) = L = \lim_{(x,y) \to (a,b)} h(x,y)$ then $\lim_{(x,y) \to (a,b)} g(x,y) = L$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $f(x,y) ≤ g(x,y) ≤ h(x,y)$ for $(x,y)$ near $(a,b)$, then $\exists \delta_0 > 0$ such that if $(x,y) \in D(f) \cap D(g) \cap D(h)$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta_0$ then $f(x,y) ≤ g(x,y) ≤ h(x,y)$.

- Now since $\lim_{(x,y) \to (a,b)} f(x,y) = L$ then $\forall \epsilon > 0$ $\exists \delta_1 > 0$ such that if $(x,y) \in D(f)$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta_0$ then $L - \epsilon ≤ f(x,y) ≤ L + \epsilon$.

- Similarly, since $\lim_{(x,y) \to (a,b)} h(x,y) = L$ then $\forall \epsilon > 0$ $\exists \delta_2 > 0$ such that if $(x,y) \in D(h)$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta_0$ then $L - \epsilon ≤ h(x,y) ≤ L + \epsilon$.

- We want to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid g(x,y) - L \mid < \epsilon$. Let $\delta = \mathrm{min} \{ \delta_0, \delta_1, \delta_2 \}$, and so we have that:

\begin{align} \quad L - \epsilon ≤ f(x,y) ≤ g(x,y) ≤ h(x,y) ≤ L + \epsilon \end{align}

- Therefore $\lim_{(x,y) \to (a,b)} g(x,y) = L$. $\blacksquare$