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

Short Answer

Expert verified
Question: Prove that the limit of the sum of two functions is equal to the sum of their limits as (x, y) approaches (a, b) using the formal definition of a limit. Answer: To prove that $\lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=\lim _{(x, y) \rightarrow(a, b)} f(x, y)+\lim _{(x, y) \rightarrow(a, b)} g(x, y)$, we showed that for any given ε > 0, there exists a δ > 0 such that if 0 < ||(x, y) - (a, b)|| < δ, then |(f(x, y)+g(x, y)) - (L1 + L2)| < ε, where L1 and L2 are the limits of f and g, respectively.

Step by step solution

01

Introduce the given problem

We are given that \(\lim _{(x, y) \rightarrow(a, b)} f(x,y) = L_1\) and \(\lim _{(x, y) \rightarrow(a, b)} g(x,y) = L_2\). We want to prove that \(\lim _{(x, y) \rightarrow(a, b)} (f(x, y) + g(x, y)) = L_1 + L_2\) using the formal definition of a limit.
02

State the formal definition of a limit

According to the formal definition of a limit, for any given ε > 0, there exists a δ > 0 such that if 0 < ||(x, y) - (a, b)|| < δ, then |(f(x, y)+g(x, y)) - (L1 + L2)| < ε.
03

Use the triangle inequality

Using the triangle inequality, we can write |(f(x, y)+g(x, y)) - (L1 + L2)| as |(f(x, y) - L1) + (g(x, y) - L2)| ≤ |f(x, y) - L1| + |g(x, y) - L2|.
04

Apply the given limits and introduce new variables

Since we know that the limits of f and g are L1 and L2, respectively, there exist δ1 and δ2 such that if 0 < ||(x, y) - (a, b)|| < δ1, then |f(x, y) - L1| < ε/2, and if 0 < ||(x, y) - (a, b)|| < δ2, then |g(x, y) - L2| < ε/2.
05

Find the δ that works for both functions

Let δ = min(δ1, δ2). Then, if 0 < ||(x, y) - (a, b)|| < δ, we have |f(x, y) - L1| < ε/2 and |g(x, y) - L2| < ε/2.
06

Combine the inequalities and complete the proof

Now, if 0 < ||(x, y) - (a, b)|| < δ, we have |(f(x, y) + g(x, y)) - (L1 + L2)| ≤ |f(x, y) - L1| + |g(x, y) - L2| < ε/2 + ε/2 = ε. Thus, we have shown that for any given ε > 0, there exists a δ > 0 such that if 0 < ||(x, y) - (a, b)|| < δ, then |(f(x, y)+g(x, y)) - (L1 + L2)| < ε, which proves that $\lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=\lim _{(x, y) \rightarrow(a, b)} f(x, y)+\lim _{(x, y) \rightarrow(a, b)} g(x, y)$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
In the realm of multivariable calculus, we explore functions that have more than one input variable. This complexity requires us to consider limits in a multi-dimensional space. For instance, when we mention the limit of a function as \( (x, y) \to (a, b) \), we're specifying that both \( x \) and \( y \) are approaching \( a \) and \( b \), respectively.

Understanding the behavior of functions at specific points, or as variables approach certain values, is key in multivariable calculus. These concepts are crucial for advanced topics such as gradient descent in optimization, calculating volumes under surfaces, and in the study of vector fields. In the given exercise, we delve into the limit of a sum of two functions, each with two variables, which is an essential skill in this field.
Limit Proof
Proving limits within multivariable calculus often requires more than intuition; it requires a formal proof using the limits' definitions and properties. The exercise presents a proof scenario involving the limit of a sum of two functions. To prove a limit assertion, we typically start by stating what we are given and what we need to show.

Subsequently, we use known theorems, properties, and definitions—such as the epsilon-delta definition of a limit—to rigorously show that our limit holds under any given criteria. In our step-by-step solution, we carefully verify that for every choice of \( \epsilon > 0 \) there exists a \( \delta > 0 \) fulfilling the necessary inequality, thus proving the assertion.
Triangle Inequality
The triangle inequality is an indispensable tool in analyzing and proving statements about limits. Simply put, it states that for any real numbers \( a \) and \( b \) the absolute value of their sum is less than or equal to the sum of their absolute values: \( |a + b| \leq |a| + |b| \).

This inequality allows us to break down complex expressions and handle them piece by piece, making it easier to work with the absolute differences required in limit definitions. In our solution, the triangle inequality is used to separate the absolute difference of the sum into the sum of absolute differences, facilitating the subsequent steps of finding appropriate \( \delta \) values.
Epsilon-Delta Definition
The epsilon-delta definition is the formal basis for limits in calculus. It provides a mathematical foundation for what it means for a function to approach a certain value as its arguments approach given points. Specifically, the definition states that the limit of \( f(x, y) \) as \( (x, y) \to (a, b) \) is \( L \) if for any \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that whenever \( 0 < ||(x, y) - (a, b)|| < \delta \) it follows that \( |f(x, y) - L| < \epsilon \).

This definition quantifies the concept of 'approaching' and 'getting arbitrarily close,' which allows us to discuss continuity and limits with rigor and precision. In our exercise, we apply this definition to prove the limit of the sum of two functions by finding a \( \delta \) that works for both functions using their individual \( \delta \) values.

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Most popular questions from this chapter

Let \(h\) be continuous for all real numbers. a. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{x}^{y} h(s) d s\). b. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{1}^{x y} h(s) d s\).

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