Problem 28 - Calculus: Early Transcendentals

Question

Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. $$\left.\lim _{x \rightarrow 3}(x-3)^{2}=0 \text { (Hint: Use the identity } \sqrt{x^{2}}=|x| .\right)$$

Step-by-Step Solution

Answer

Question: Prove that the limit of \((x-3)^2\) as \(x\) approaches \(3\) is \(0\) using the precise definition of a limit and the identity \(\sqrt{x^2} = |x|\). Answer: We showed that, for any given \(\varepsilon > 0\), we can choose \(\delta = \sqrt[3]{\varepsilon}\) such that if \(0<|x-3|<\delta\), then \(|(x-3)^2|<\varepsilon\). This satisfies the precise definition of a limit, proving that \(\lim_{x \rightarrow 3} (x-3)^2 = 0\).

Step 1: The precise definition of a limit states that for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 0\), we can find a suitable \(\delta > 0\) that satisfies the condition. #Step 2: Working with the absolute value function#

To prove the limit, we need to manipulate the equation \(|f(x)-L|=|x-3|^2\). In this case, our goal is to connect this function with the distance between \(x\) and \(a\), which is \(|x-3|\). We will use the provided hint: remember that \(\sqrt{x^2} = |x|\). So, we can rewrite our equation as follows: $$|x-3|^2 = |\sqrt{(x-3)^4}|$$ #Step 3: Introducing epsilon and delta#

Step 2: We want to show that \(|(x-3)^4| < \varepsilon\) when \(0 < |x-3| < \delta\). Notice that \((x-3)^4\) is a monotonically increasing function – that is, as \(x\) gets larger, so does \((x-3)^4\). From this, we can intuitively guess that we can choose \(\delta\) to be the positive cubic root of \(\varepsilon\), or \(\delta = \sqrt[3]{\varepsilon}\). #Step 4: Proving the limit exists#

We will now prove that our choice of \(\delta = \sqrt[3]{\varepsilon}\) ensures the limit exists. If \(0 < |x-3| < \delta\), then: $$0 < |x-3|^4 < \delta^4$$ Due to our choice of \(\delta\), we have: $$0 < |x-3|^4 < \varepsilon$$ Thus, for any given \(\varepsilon > 0\), we can choose \(\delta = \sqrt[3]{\varepsilon}\) such that if \(0 < |x-3| < \delta\), then \(|(x-3)^2| < \varepsilon\). This satisfies the precise definition of a limit and therefore proves that: $$\lim _{x \rightarrow 3} (x-3)^2 = 0$$

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a mathematical tool used to formally define the limit of a function. It states that for a limit to exist at a point, you must be able to make the function values arbitrary close to a target value (the limit) by choosing input values that are sufficiently close to a specified number.
In formal terms, for a function \( f(x) \) approaching a limit \( L \) as \( x \) approaches \( a \), for every positive number \( \varepsilon \) (representing how close we want \( f(x) \) to be to \( L \)), there must be a positive number \( \delta \) such that whenever \( 0 < |x-a| < \delta \), it will follow \( |f(x)-L| < \varepsilon \).

• \( \varepsilon \) is how close we want \( f(x) \) to be to \( L \).
• \( \delta \) is how close \( x \) needs to be to \( a \) to meet our \( \varepsilon \) criteria.

This precise definition ensures that the limit holds rigorous control over function behavior as \( x \) approaches a given value, verifying limit existence beyond visual interpretations.

Monotonic Functions

Monotonic functions are those that consistently increase or decrease across their domain. They are key in limit proofs due to their predictability. Understanding how they behave is crucial in determining the right \( \delta \) that correlates with \( \varepsilon \).
In our exercise, the expression \( (x-3)^4 \) is a monotonic function because as \( x \) increases or decreases, \( (x-3)^4 \) consistently increases.

• A function is increasing if for every \( x_1 < x_2 \), we have \( f(x_1) \leq f(x_2) \).
• Decreasing if for every \( x_1 < x_2 \), \( f(x_1) \geq f(x_2) \).

The continuous nature of monotonic functions assists in the assumption that if we pick any small \( \delta \), the corresponding function changes predictably, allowing us to establish the necessary \( \delta \) for proofing limits.

Absolute Value

The absolute value function measures the magnitude of a real number irrespective of its sign. In calculus, it plays a substantial role in limit proofs and other operations requiring distance measurements.
Mathematically, \( |x| = x \) if \( x \geq 0 \), and \( |x| = -x \) if \( x < 0 \). This definition ensures that irrespective of whether a number is positive or negative, it gives the true 'distance' from zero.
Within our proof scenario, observing that \( |x-3|^2 \) and \( |(x-3)^4| \) are absolute values allows you to linearly map function behavior concerning \( \varepsilon \) and \( \delta \). By leveraging properties like \( |a - b| \) representing the distance between \( a \) and \( b \), one can understand their role in bridging the gap between input-output relations in function limit assessments.

Cubic Root

The cubic root function, denoted as \( \sqrt[3]{x} \), is a special mathematical operation that pinpoints a number which when cubed, returns \( x \). In limit proofs and calculations, it acts as a valuable tool for transforming relations to find manageable \( \delta \) values.
In our step-by-step solution, picking \( \delta = \sqrt[3]{\varepsilon} \) relates to simplifying and managing inequalities, ensuring \( |x-3|^4 \) remains smaller than \( \varepsilon \).

• This specific example allows \( |x-3| \) transformations into ranges manageable for concrete epsilon-delta connections.
• Utilizing cubic roots helps simplify expressions resulting from higher-order polynomials, granting precise control in limit estimations.

Understanding cubic roots in the context of limits offers a method to keep expressions in check, validating them for the epsilon-delta framework.