Limit Laws for Functions of Several Variables Examples 1

# Limit Laws for Functions of Several Variables Examples 1

Recall from the Limit Laws for Functions of Several Variables page that if $\lim_{(x, y) \to (a, b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$ then:

• 1) $\lim_{(x, y) \to (a, b)} [f(x, y) + g(x, y)] = L + M$.
• 2) $\lim_{(x, y) \to (a, b)} [f(x, y) - g(x, y)] = L - M$.
• 3) $\lim_{(x, y) \to (a, b)} [f(x, y) g(x, y)] = LM$.
• 4) $\lim_{(x, y) \to (a, b)} \frac{f(x, y)}{g(x, y)} = \frac{L}{M}$ provided that $M \neq 0$.
• 5) $\lim_{(x, y) \to (a, b)} kf(x, y) = kL$ for all $k \in \mathbb{R}$.

Furthermore, we have the Squeeze theorem for functions of several variables:

• 6) If $f(x,y) ≤ g(x,y) ≤ h(x,y)$ for $(x, y)$ near $(a, b)$ and $\lim_{(x, y) \to (a, b)} f(x, y) = L = \lim_{(x, y) \to (a, b)} h(x, y)$ then $\lim_{(x, y) \to (a, b)} g(x, y) = L$.

We will now look at some examples regarding these limit laws.

## Example 1

Prove the second limit law from above.

Let $\lim_{(x, y) \to (a, b)} f(x, y) = L$ and $\lim_{(x, y) \to (a, b)} g(x, y) = M$. We want to prove that:

(1)
\begin{align} \quad \lim_{(x, y) \to (a, b)} [f(x, y) - g(x, y)] = L - M \end{align}

That is, we want to show that $\forall \epsilon > 0$ there exists a $\delta > 0$ 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 $\epsilon > 0$ be given.

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

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

Let $\delta = \min \{ \delta_1, \delta_2 \}$. Therefore by applying the triangle inequality we have that:

(2)
\begin{align} \quad \quad \mid [f(x, y) - g(x, y)] - (L - M) \mid = \mid f(x, y) - L - g(x, y) + 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 the $\delta$ prescribed above is suitable. Thus, $\lim_{(x, y) \to (a, b)} [f(x, y) - g(x, y)] = L - M$.

## Example 2

Prove the fifth limit law from above.

Let $\lim_{(x, y) \to (a, b)} f(x, y) = L$ and let $k \in \mathbb{R}$. We want to show that:

(3)
\begin{align} \quad \lim_{(x, y) \to (a, b)} kf(x, y) = kL \end{align}

We need to show that $\forall \epsilon > 0$ there exists a $\delta > 0$ such that if $(x, y) \in D(kf)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid kf(x, y) - kL \mid < \epsilon$.

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

Let $\delta = \delta_1$. Then we have that:

(4)
\begin{align} \quad \mid k f(x, y) - kL \mid = \mid k(f(x, y) - L) \mid = \mid k \mid \mid f(x, y) - L \mid < \mid k \mid \epsilon_1 = \mid k \mid \frac{\epsilon}{\mid k \mid} = \epsilon \end{align}

Therefore this $\delta$ will do, and $\lim_{(x, y) \to (a, b)} kf(x, y) = k L$.

Now suppose that $k = 0$. Then $k f(x, y) = 0$. Clearly, $\lim_{(x, y) \to (a, b)} 0 = 0$. Furthermore, $kL = 0$, so the fifth limit law still holds for $k = 0$.

Note that we could not have included this case in the formal proof above because we would then be dividing by $0$.