Question

Use the formal definition of a limit to prove that $$\begin{aligned} \lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=& \lim _{(x, y) \rightarrow(a, b)} f(x, y) \\\\+& \lim _{(x, y) \rightarrow(a, b)} g(x, y) \end{aligned}$$

Short answer

Based on the step by step solution provided, answer the following question: Question: Prove that the limit of the sum of two functions, f(x, y) and g(x, y), is equal to the sum of their limits as (x, y) approaches some point (a, b). Answer: To prove this, we first state the formal definition of a limit for the given functions and then use the Triangle Inequality to relate the sum of the functions' limits to their individual limits. Finally, we determine a suitable δ that proves the limit for the sum of the functions. By showing that |(f(x, y) + g(x, y)) - (L_f + L_g)| < ε for the chosen δ, we prove that the limit of the sum of the functions is equal to the sum of their limits as (x, y) approaches (a, b).

Step-by-step solution

State the Formal Definition of a Limit for the Given Functions

By the formal definition of a limit, given ε_f > 0 and ε_g > 0, there exist δ_f > 0 and δ_g > 0 such that if 0 < |(x, y) - (a, b)| < δ_f, then |f(x, y) - L_f| < ε_f, and if 0 < |(x, y) - (a, b)| < δ_g, then |g(x, y) - L_g| < ε_g. Now our goal is to find a δ > 0 such that if 0 < |(x, y) - (a, b)| < δ, then |(f(x, y) + g(x, y)) - (L_f + L_g)| < ε.

Utilize the Triangle Inequality

By the Triangle Inequality, we have: |(f(x, y) + g(x, y)) - (L_f + L_g)| ≤ |f(x, y) - L_f| + |g(x, y) - L_g| Let's analyze the right-hand side of the equation. Since 0 < |(x, y) - (a, b)| < δ_f, we have |f(x, y) - L_f| < ε_f. Similarly, 0 < |(x, y) - (a, b)| < δ_g, so |g(x, y) - L_g| < ε_g. Thus, we can conclude that: |(f(x, y) + g(x, y)) - (L_f + L_g)| ≤ ε_f + ε_g

Determine a Suitable δ

To prove that the limit of the sum of the functions exists and is equal to the sum of their limits, let's take ε = ε_f + ε_g and choose δ = min(δ_f, δ_g). Since both δ_f and δ_g exist, we have δ > 0. Now, if 0 < |(x, y) - (a, b)| < δ, then |(x, y) - (a, b)| < δ_f and |(x, y) - (a, b)| < δ_g. By our analysis in Step 2, this implies that |(f(x, y) + g(x, y)) - (L_f + L_g)| < ε_f + ε_g = ε. Therefore, by the formal definition of a limit, we have: $$\begin{aligned} \lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=& \lim _{(x, y) \rightarrow(a, b)} f(x, y) \\\\+& \lim _{(x, y) \rightarrow(a, b)} g(x, y) \end{aligned}$$ and the proof is complete.

Key concepts

Formal definition of limit

Understanding the formal definition of a limit is crucial in proving properties of functions like their sum, product, or composition. The formal definition states that a function \( f(x, y) \) approaches a limit \( L_f \) as the point \( (x, y) \) approaches \( (a, b) \) if for every positive number \( \varepsilon_f \), no matter how small, there exists a positive number \( \delta_f \) such that whenever the distance from \( (x, y) \) to \( (a, b) \) is less than \( \delta_f \), then the absolute difference between \( f(x, y) \) and \( L_f \) is less than \( \varepsilon_f \).

This sounds complicated, but let's break it down. Essentially, it means that as you get closer to the point \( (a, b) \), \( f(x, y) \) gets closer to \( L_f \). You can tighten how close \( f(x, y) \) must get to \( L_f \) (by choosing a smaller \( \varepsilon_f \)), and there will always be a small neighborhood around \( (a, b) \) (determined by \( \delta_f \)) where this proximity is assured.

Triangle inequality

The triangle inequality is a fundamental concept in mathematics that provides a relationship between the lengths of sides of a triangle. In the context of limits and functions, it gives us a tool to approximate the behavior of sums or differences of functions.

Mathematically, the triangle inequality is stated as \( |u + v| \leq |u| + |v| \). This means that the absolute value of the sum of two numbers \( u \) and \( v \) is always less than or equal to the sum of the absolute values of \( u \) and \( v \).

In our limit problem, we apply the triangle inequality to the expression \( |(f(x, y) + g(x, y)) - (L_f + L_g)| \). This becomes \( |f(x, y) - L_f| + |g(x, y) - L_g| \), allowing us to handle the components \( |f(x, y) - L_f| \) and \( |g(x, y) - L_g| \) separately. These components are less than \( \varepsilon_f \) and \( \varepsilon_g \) respectively, provided \( \delta_f \) and \( \delta_g \) are chosen appropriately.

Limit of sum of functions

The concept of the limit of the sum of functions allows us to deduce the limit of combined functions based on their individual limits. If you know that \( \lim_{(x, y) \to (a, b)} f(x, y) = L_f \) and \( \lim_{(x, y) \to (a, b)} g(x, y) = L_g \), then the limit of their sum is \( \lim_{(x, y) \to (a, b)} (f(x, y) + g(x, y)) = L_f + L_g \).

Using the triangle inequality, we establish that \( |(f(x, y) + g(x, y)) - (L_f + L_g)| \leq |f(x, y) - L_f| + |g(x, y) - L_g| \). For the limit to hold, both components must be less than their respective epsilon values \( \varepsilon_f \) and \( \varepsilon_g \).

To prove the limit of the sum, select a \( \delta \) as the minimum of \( \delta_f \) and \( \delta_g \). This \( \delta \) assures that \( (x, y) \) is sufficiently close to \( (a, b) \) so that \( |f(x, y) - L_f| < \varepsilon_f \) and \( |g(x, y) - L_g| < \varepsilon_g \). Therefore, \( |(f(x, y) + g(x, y)) - (L_f + L_g)| < \varepsilon_f + \varepsilon_g = \varepsilon \), confirming that the sum of the limits equals the limit of the sum.