Problem 25 - Calculus: Early Transcendentals

Question

Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to 0} x^2 = 0 \)

Step-by-Step Solution

Answer

Limit is proven using \( \varepsilon, \delta \) for any \( \varepsilon > 0 \) with \( \delta = \sqrt{\varepsilon} \).

Step 1: Understanding the Limit Definition

The definition of a limit states that \( \lim_{{x \to c}} f(x) = L \) if for every \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \varepsilon \). In this case, \( f(x) = x^2 \), \( c = 0 \), and \( L = 0 \). So, we need to find \( \delta \) such that if \( 0 < |x| < \delta \), then \( |x^2 - 0| < \varepsilon \).

Step 2: Expressing the Function

Given the function \( f(x) = x^2 \), we want \( |x^2 - 0| < \varepsilon \). This simplifies to \( |x^2| < \varepsilon \), which further simplifies to \( x^2 < \varepsilon \).

Step 3: Relating \( x^2 \) to \( |x| \)

We rewrite the inequality \( x^2 < \varepsilon \) to find a condition on \( |x| \): Since \( x^2 < \varepsilon \) implies \( |x| < \sqrt{\varepsilon} \), we know that satisfying \( |x| < \sqrt{\varepsilon} \) also ensures \( x^2 < \varepsilon \).

Step 4: Choosing \( \delta \)

To satisfy the limit definition, we choose \( \delta = \sqrt{\varepsilon} \). This choice means that whenever \( 0 < |x| < \delta \), we have \( 0 < |x| < \sqrt{\varepsilon} \), which guarantees \( x^2 < \varepsilon \).

Step 5: Verifying the Solution

Now, if \( 0 < |x| < \delta \), then \( 0 < |x| < \sqrt{\varepsilon} \), leading to \( x^2 < \varepsilon \), thus ensuring \( |f(x) - L| = |x^2 - 0| < \varepsilon \). This successfully confirms that the limit \( \lim_{{x \to 0}} x^2 = 0 \) is validated by the \( \varepsilon \), \( \delta \) definition.

Key Concepts

Limits in Calculus

In calculus, the concept of a limit explores the behavior of a function as its input approaches a certain point. This is a fundamental aspect, acting as the foundation for more advanced topics like derivatives and integrals. When we say the limit of a function \( f(x) \) as \( x \) approaches a value \( c \) is \( L \), it means that as \( x \) gets closer to \( c \), the function \( f(x) \) gets closer to \( L \).
In our example, \( \lim_{x \to 0} x^2 = 0 \) expresses that as \( x \) approaches 0, the value of \( x^2 \) approaches 0 as well. This idea captures not just what happens exactly at a point, but rather the trend of approaching values. It's particularly useful when dealing with points where the function is undefined or unpredictable.
Limits help us understand the continuity and predictability of mathematical functions, which is pivotal in calculus.

Continuity

Continuity in mathematics refers to a function that has no interruptions, jumps, or holes in its graph. For a function to be continuous at a point, the limit of the function as it approaches the point must equal the function's value at that point. This is captured by three conditions:

• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The function \( f(x) \) is defined at \( c \), meaning \( f(c) \) is a real number.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

When dealing with the parabola represented by \( x^2 \), the function is continuous everywhere because it has a smooth and unbroken curve. Specifically, at \( x = 0 \), \( x^2 \) meets all these criteria for continuity since \( \lim_{x \to 0} x^2 = 0 \) is equal to \( (0)^2 = 0 \). This continuous nature makes functions much easier to work with in calculus as it allows the use of limits and derivations confidently without worrying about discontinuities.

Epsilon-Delta Proof

The \( \varepsilon \)-\( \delta \) definition of a limit is a rigorous way to prove that the limit of a function exists and is equal to a particular value. Here's how the method works using the example \( \lim_{x \to 0} x^2 = 0 \):

• First, we're given \( \varepsilon > 0 \) and need to find a corresponding \( \delta > 0 \) such that whenever \( 0 < |x| < \delta \), we have \( |x^2 - 0| < \varepsilon \).
• For the function \( f(x) = x^2 \), the task simplifies to ensuring \( x^2 < \varepsilon \), which implies that \( |x| < \sqrt{\varepsilon} \).
• Thus, choosing \( \delta = \sqrt{\varepsilon} \) guarantees that \( x^2 < \varepsilon \) whenever \( 0 < |x| < \sqrt{\varepsilon} \), thus satisfying the \( \varepsilon \)-\( \delta \) condition.

The proof validates that no matter how small \( \varepsilon \) is, a suitable \( \delta \) can always be found to maintain \( |x^2| < \varepsilon \). This rigor forms the backbone of calculus, offering a clear and precise way to prove limits, crucial for ensuring function behavior stays predictable and stable.