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)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)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)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)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 $0$.