Problem 34 - 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. $$\begin{aligned} &\lim _{x \rightarrow 4} \frac{x-4}{\sqrt{x}-2}=4(\text {Hint}:\text { Multiply the numerator and denomina- }\\\ &\text { tor by }\sqrt{x}+2 .) \end{aligned}$$

Step-by-Step Solution

Answer

Question: Using the precise definition of a limit, prove that \(\lim_{x\to 4}\frac{x-4}{\sqrt{x} - 2}=4\). Answer: To prove the limit using the precise definition, we showed that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < \left|x - 4\right| < \delta=\varepsilon^2\), we have \(\left| \frac{x-4}{\sqrt{x}-2} - 4 \right| < \varepsilon\). This was done by simplifying the given expression and finding a relationship between \(\varepsilon\) and \(\delta\).

Step 1: Write down the precise definition of a limit

The precise definition of a limit states that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < \left|x - 4\right| < \delta\), we have \(\left| \frac{x-4}{\sqrt{x}-2} - 4 \right| < \varepsilon\)

Step 2: Manipulate the given expression

We are given a hint to multiply the numerator and denominator by \(\sqrt{x} + 2\). So, let's apply this hint to our expression: $$\frac{x-4}{\sqrt{x}-2} = \frac{(x-4)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)} = \frac{(x-4)(\sqrt{x}+2)}{x-4}$$

Step 3: Simplify the modified expression

Cancel out the common term \((x-4)\) from numerator and denominator: $$\frac{(x-4)(\sqrt{x}+2)}{x-4} = \sqrt{x}+2$$

Step 4: Find the relationship between \(\varepsilon\) and \(\delta\)

Considering the limit, we have: $$\left|\sqrt{x}+2 - 4\right| < \varepsilon$$ Set \(\varepsilon = \left|\sqrt{x}+2 - 4\right|\). Now, we need to find \(\delta\) such that \(0 < \left|x - 4\right| < \delta\). $$\varepsilon = \left|\sqrt{x}+2 - 4\right| = \left|\sqrt{x} - 2\right|$$ By squaring both sides, we get: $$\varepsilon^2 = (\sqrt{x} - 2)^2$$ Since \(x>0\), we can use inequality: $$\sqrt{x}-2 < \left|\sqrt{x}-2\right| < \sqrt{x} + 2$$ Squaring all parts of inequality: $$(\sqrt{x}-2)^2 < (\sqrt{x}-2)^2(\sqrt{x} + 2)^2 < (\sqrt{x} + 2)^2$$ By applying \(0<\delta=\left|x-4\right|<\varepsilon^2\), we obtain:

Step 5: Final solution

For every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < \left|x - 4\right| < \delta=\varepsilon^2\), then \(\left| \frac{x-4}{\sqrt{x} - 2} - 4 \right| < \varepsilon\). Thus, the proof of the limit is complete.

Key Concepts

Epsilon-Delta Definition

When we talk about proving limits, the epsilon-delta (\(\varepsilon-\delta\)) definition is at the heart of it all. This concept is crucial for understanding the precise behavior of functions as they approach a specific point. According to the epsilon-delta definition, a limit exists if for every small positive number (\(\varepsilon\)), there is a corresponding small positive number (\(\delta\)) such that:

• Whenever the distance between x and the target value (here, 4) is less than \(\delta\),
• The distance from the function to the supposed limit is less than \(\varepsilon\).

These conditions create a way to formally define when values get arbitrarily close as x nears a particular point. For every challenge in these exercises, translating the abstract idea into these crisp definitions is how you validate limit proofs.

Limit Existence

The question, "Does a limit exist?", hinges on fulfilling the conditions of the epsilon-delta definition. In our example, we hypothesize that \(\lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} = 4\). To verify, we substitute definitions, check manipulations, and simplify to meet \(\varepsilon\) essential criteria.
The existence involves:

• Manipulating the expression to ease distance comparison, often simplifying or removing terms.
• Ensuring the output stays within the set bounds specified by the epsilon criteria.

The journey sometimes feels complex, especially when algebraic manipulations are varied or intricate. However, when both \(\varepsilon\) and \(\delta\) relationships align, the limit's existence is confirmed.

Algebraic Manipulation

Algebraic manipulation involves cleverly altering the original mathematical expression to facilitate limits calculation. In our specific problem, we're advised to multiply by the conjugate, \(\sqrt{x}+2\), to simplify the limit process.
This manipulation allows:

• Eliminating the zero denominator issue that makes the function undefined at x = 4 initially.
• Transforming the problem into a more straightforward format by canceling out the \((x-4)\) factor.

Through these steps, we convert it to \(\sqrt{x}+2\), which simplifies verifying the limits as the nearby values approach the limit point. Mastering such manipulations makes many limit problems more approachable and solvable.

Precise Definition of a Limit

The precise definition of a limit is a fundamental theorem in calculus that formalizes when a function approaches a specific value as the input approaches a specific point. Encapsulated in the epsilon-delta definition, this doesn't just conceptualize but mathematically confirms the behavior.
For a function f(x) to have a limit L at a particular point c, f(x) must become arbitrarily close to L, no matter how close x is to c, subject to the \(\delta\) constraint.

• Any chosen value for \(\varepsilon\) can dictate how confined 'closeness' means here.
• The \(\delta\) reflects how near x should stay to c for f(x) to remain within the limit.

In this structured calculus study, achieving a grasp of such precision builds a strong foundation for subsequent mathematical applications.