Problem 9 - Calculus for Biology and Medicine

Question

Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow 2}(2 x-1)=3. $$

Step-by-Step Solution

Answer

Using \( \delta = \frac{\epsilon}{2} \), the limit \( \lim _{x \to 2} (2x - 1) = 3 \) is proven by the epsilon-delta definition.

Step 1: Understanding the Definition of a Limit

Recall the formal (epsilon-delta) definition of a limit: \( \lim_{x \to a} f(x) = L \) means that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \). In this problem, \( a = 2 \), \( L = 3 \), and \( f(x) = 2x - 1 \).

Step 2: Applying the Definition

We need to show for every \( \epsilon > 0 \), there is a \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \) then \( |(2x - 1) - 3| < \epsilon \). Simplifying \( |(2x - 1) - 3| \), we get \( |2x - 4| \), which further simplifies to \( 2|x - 2| \).

Step 3: Determining \( \delta \)

Now, we have the inequality \( 2|x - 2| < \epsilon \). To satisfy this inequality, we choose \( |x - 2| < \frac{\epsilon}{2} \). Therefore, we set \( \delta = \frac{\epsilon}{2} \).

Step 4: Conclusion of the Proof

With \( \delta = \frac{\epsilon}{2} \), the condition \( 0 < |x - 2| < \delta \) implies \( 2|x - 2| < \epsilon \), which satisfies the formal definition of limits. Thus, using the epsilon-delta definition, we have verified that \( \lim _{x \rightarrow 2}(2 x-1)=3 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition of a limit is a fundamental part of calculus that explains what it means for a function to approach a specific value as its input approaches a given point. More formally, when we say \( \lim_{{x \to a}} f(x) = L \), we are stating that for every small number \( \epsilon > 0 \), there is a corresponding small distance \( \delta > 0 \) from the point \( a \), within which the value of \( f(x) \) remains within \( \epsilon \) of \( L \).

• Here, \( \epsilon \) represents the closeness of \( f(x) \) to \( L \).
• \( \delta \) represents the distance you can move from \( a \) such that all \( f(x) \) values meet the \( \epsilon \) closeness requirement.

Let's break this down: for every \( \epsilon \), you can find a \( \delta \) where if \( x \) is within \( \delta \) of \( a \) (but \( x eq a \)), then \( f(x) \) is within \( \epsilon \) of \( L \). This concept is crucial to understanding limits as it provides a rigorous framework to prove a function's limit at a point.

Calculus Proof

In calculus, proving something can require us to rigorously show how definitions translate into mathematical reasoning. Let's see how this works with the epsilon-delta definition by proving that \( \lim_{x \rightarrow 2}(2x-1)=3 \).

• Identify the function \( f(x) = 2x-1 \), point \( a = 2 \), and the limit \( L = 3 \).
• Simplify the expression \(|f(x)-L|\) to \(|(2x-1) - 3| = |2x - 4| = 2|x-2|\).

The goal is to find \( \delta \) for each \( \epsilon \) such that \( 0 < |x-2| < \delta \) leads to \( |2x-4| < \epsilon \). Simplifying gives the condition \( 2|x-2| < \epsilon \). Therefore, if we let \( |x-2| < \frac{\epsilon}{2} \), any \( \epsilon \) leads us to choose \( \delta = \frac{\epsilon}{2} \).By finding this \( \delta \), we ensure that the original inequality condition is satisfied. Thus, using the epsilon-delta definition, we verify that the limit statement holds true.

Continuity

A function is continuous at a point \( a \) if the following three conditions are met:

• \( f(a) \) is defined. This means that when you plug in \( a \) into the function, you get a specific value.
• \( \lim_{x \to a} f(x) \) exists. The function approaches a definite value as \( x \) approaches \( a \).
• \( \lim_{x \to a} f(x) = f(a) \). The approaching value and the actual value at \( a \) are the same.

In simpler terms, for a function to be continuous at a point, it cannot "jump" or "break" at that point.
Consider our example with the function \( f(x) = 2x-1 \). At \( x = 2 \), the function is continuous because:

• It is defined at \( x = 2 \), giving \( f(2) = 3 \).
• The limit as \( x \) approaches 2 is also 3 (as shown in the proof).
• The limit matches the function value at \( x = 2 \), maintaining continuity.

Understanding continuity helps explain why functions behave nicely and predictably at certain points, a key concept when tackling calculus problems.