Problem 6 - Calculus

Question

Suppose \(f(x) \rightarrow L\) and \(g(x) \rightarrow M\) as \(x \rightarrow a\). Prove from the definitions that \(f(x)+g(x) \rightarrow L+M\) as \(x \rightarrow a\)

Step-by-Step Solution

Answer

The sum \(f(x) + g(x)\) approaches \(L + M\) as \(x\) approaches \(a\).

Step 1: Understanding the Problem

We need to prove that if the functions \(f(x)\) approaches \(L\) and \(g(x)\) approaches \(M\) as \(x\) approaches \(a\), then the sum \(f(x) + g(x)\) approaches \(L + M\) as \(x\) approaches \(a\). This involves using the definition of a limit.

Step 2: Expressing Limits with Definitions

According to the definition of limits, \(f(x) \rightarrow L\) as \(x \rightarrow a\) means that for every \(\epsilon_1 > 0\), there exists a \(\delta_1 > 0\) such that if \(0 < |x - a| < \delta_1\), then \(|f(x) - L| < \epsilon_1\). Similarly, \(g(x) \rightarrow M\) as \(x \rightarrow a\) means for every \(\epsilon_2 > 0\), there is a \(\delta_2 > 0\) such that if \(0 < |x - a| < \delta_2\), then \(|g(x) - M| < \epsilon_2\).

Step 3: Combining the Limits

We need to show that \(f(x) + g(x) \rightarrow L + M\) as \(x \rightarrow a\). For this, let \(\epsilon > 0\) be given. We need to find a \(\delta > 0\) such that if \(0 < |x - a| < \delta\), then \(|(f(x) + g(x)) - (L + M)| < \epsilon\).

Step 4: Applying the Triangle Inequality

Using the triangle inequality, \(|(f(x) + g(x)) - (L + M)| = |(f(x) - L) + (g(x) - M)| \leq |f(x) - L| + |g(x) - M|\).

Step 5: Choosing the Appropriate Delta

To make \(|f(x) - L| + |g(x) - M| < \epsilon\), we can choose \(\epsilon_1 = \epsilon/2\) and \(\epsilon_2 = \epsilon/2\). Then, there exists \(\delta_1 > 0\) such that for \(0 < |x - a| < \delta_1\), we have \(|f(x) - L| < \epsilon/2\), and \(\delta_2 > 0\) such that for \(0 < |x - a| < \delta_2\), we have \(|g(x) - M| < \epsilon/2\).

Step 6: Finalizing the Proof

Let \(\delta = \min(\delta_1, \delta_2)\). Then, for \(0 < |x - a| < \delta\), we have both \(|f(x) - L| < \epsilon/2\) and \(|g(x) - M| < \epsilon/2\). Hence, \(|(f(x) + g(x)) - (L + M)| \leq |f(x) - L| + |g(x) - M| < \epsilon/2 + \epsilon/2 = \epsilon\). Thus, we have shown that \(f(x) + g(x) \rightarrow L + M\) as \(x \rightarrow a\).

Key Concepts

Definition of Limit

Understanding the concept of limits is essential in calculus. In simple terms, when we say that the limit of a function \(f(x)\) as \(x\) approaches a value \(a\) is \(L\), it implies that as \(x\) gets closer to \(a\), \(f(x)\) gets closer to \(L\).
For every number \(\epsilon > 0\), there is a corresponding number \(\delta > 0\) such that for all \(x\) within distance \(\delta\) from \(a\) (but not equal to \(a\)), the value of \(f(x)\) remains within distance \(\epsilon\) from \(L\).
This can be mathematically stated as:

• If \(0 < |x - a| < \delta\), then \(|f(x) - L| < \epsilon\).

This precise definition allows us to pin down the behavior of functions close to a particular point and is a foundational element of calculus.

Triangle Inequality

The triangle inequality is a fundamental concept in mathematics, particularly in the context of distances in real space. 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.
This can be mathematically expressed as:

• \(|a + b| \leq |a| + |b|\)

In the context of limits and calculus, the triangle inequality helps us manage the potential sum of errors when combining two functions. For the limit involving \(f(x) + g(x)\), the triangle inequality allows us to estimate the distance of \((f(x) + g(x))\) from \((L + M)\) as not exceeding the sum of the distances of \(f(x)\) from \(L\) and \(g(x)\) from \(M\).
Using this principle, we can break complex relationships into simpler parts, making challenges easier to solve.

Delta-Epsilon Proof

The delta-epsilon (\(\delta-\epsilon\)) proof is a crucial technique to rigorously prove limits. It involves demonstrating that the function \(f(x)\) can be made arbitrarily close to a proposed limit \(L\) by making \(x\) sufficiently close to \(a\).
Here's how the delta-epsilon proof conceptually works:

• Given \(\epsilon > 0\), determine a \(\delta > 0\) such that for all \(x\), if \(0 < |x - a| < \delta\), then \(|f(x) - L| < \epsilon\).

This proof method works by establishing a direct link between the distance \(\epsilon\) from the limit \(L\) and the distance \(\delta\) from the point \(a\).
For the case \(f(x) + g(x) \rightarrow L + M\), we used this process:
- Assigned \(\epsilon/2\) to each function’s distance requirement.
- Chose \(\delta\) as the smaller of the two deltas needed for each function. This ensures both conditions \(|f(x) - L| < \epsilon/2\) and \(|g(x) - M| < \epsilon/2\) hold simultaneously, guaranteeing \(|(f(x) + g(x)) - (L + M)| < \epsilon\).
This approach, although sometimes intricate, is the basis for proving many key results in calculus with precision.