Problem 62 - Calculus Early Transcendentals

Question

$$ \begin{array}{l}{\text { For the limit }} \\ {\quad \lim _{x \rightarrow \infty} \frac{\sqrt{4 x^{2}+1}}{x+1}=2} \\ {\text { illustrate Definition } 7 \text { by finding values of } N \text { that correspond }} \\ {\text { to } \varepsilon=0.5 \text { and } \varepsilon=0.1}\end{array} $$

Step-by-Step Solution

Answer

For \( \varepsilon=0.5 \), choose \( N = 2 \); for \( \varepsilon=0.1 \), choose \( N = 10 \).

Step 1: Understand the Limit Problem

We are given the limit \( \lim_{x \to \infty} \frac{\sqrt{4x^2+1}}{x+1} = 2 \). According to the problem, we need to illustrate this limit using a formal definition of a limit related to \( \varepsilon \), and find \( N \) for \( \varepsilon = 0.5 \) and \( \varepsilon = 0.1 \).

Step 2: Recall the Definition of the Limit at Infinity

The definition of the limit \( \lim_{x \to \infty} f(x) = L \) using \( \varepsilon \) states that for every \( \varepsilon > 0 \), there exists a number \( N \) such that for all \( x > N \), \( |f(x) - L| < \varepsilon \).

Step 3: Set Up the Inequality for \( \varepsilon = 0.5 \)

We need to find \( N \) such that for \( x > N \), \( \left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| < 0.5 \).

Step 4: Simplify the Inequality for \( \varepsilon = 0.5 \)

Start simplifying: \( \left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| < 0.5 \).Express the difference:\[\frac{\sqrt{4x^2+1} - 2(x+1)}{x+1} = \frac{\sqrt{4x^2+1} - 2x - 2}{x+1}\]This inequality becomes:\[ \left| \frac{\sqrt{4x^2+1} - 2x - 2}{x+1} \right| < 0.5 \].

Step 5: Analyze and Find Relationship

Since \( \sqrt{4x^2+1} \approx 2x \) as \( x \to \infty \), we want the numerator \( \sqrt{4x^2+1} - 2x - 2 \) to be less than \( 0.5(x+1) \) in absolute value for a sufficiently large \( x \).

Step 6: Determine \( N \) for \( \varepsilon = 0.5 \)

Solve:\[\left| \frac{\sqrt{4x^2+1} - 2x - 2}{x+1} \right| < 0.5 \]Numerator \( \approx 1 \), so:\[\frac{1}{x+1} < 0.5 \]Therefore, \( x+1 > 2 \), hence \( x > 1 \). Choose \( N = 2 \).

Step 7: Set Up the Inequality for \( \varepsilon = 0.1 \)

We repeat the steps for \( \varepsilon = 0.1 \), solving \( \left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| < 0.1 \).

Step 8: Determine \( N \) for \( \varepsilon = 0.1 \)

Numerator \( \approx 1 \), then solve:\[\frac{1}{x+1} < 0.1 \]Hence, \( x+1 > 10 \), so \( x > 9 \). Choose \( N = 10 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition of limits is a formal way to describe how a function behaves as it approaches a particular point or infinity. Consider the statement \( \lim_{x \to \infty} f(x) = L \). This means that for every small positive number \( \varepsilon \), there is a corresponding large number \( N \) such that if \( x > N \), the absolute difference between \( f(x) \) and \( L \) is less than \( \varepsilon \). This is expressed mathematically as:

• For every \( \varepsilon > 0 \), there exists an \( N \) such that \( |f(x) - L| < \varepsilon \) for all \( x > N \).

This definition captures the idea that as \( x \) gets very large, \( f(x) \) gets arbitrarily close to \( L \). In practice, it helps us find how far out we need to "zoom" along the x-axis before the function remains within a given distance \( \varepsilon \) of \( L \). This demystifies how limits at infinity are not just about approaching a number near a point but about a predictable pattern as values grow.

Numerical Limits

Numerical limits often involve finding specific values for functions as they tend toward infinity or particular points. This usually means using numeric calculations to understand a function's behavior at extreme values. In our problem, we calculated numerical limits by finding \( N \) values for \( \varepsilon = 0.5 \) and \( \varepsilon = 0.1 \). For \( \varepsilon = 0.5 \):

• The inequality \( \left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| < 0.5 \) translates to finding where \( x > 1 \), suggesting an \( N \) of 2.

On the other hand, when \( \varepsilon = 0.1 \), the same process tells us \( x > 9 \), resulting in \( N = 10 \). This approach provides concrete values for what seems like abstract mathematical behavior. Breaking down the components into solvable equations helps to visually understand how the function behaves.

Calculus Problems

Calculus problems often involve finding limits, derivatives, and integrals. These problems can be solved by applying specific theorems and formulas to understand complex behaviors. In the limit problem we studied, it was crucial to apply the epsilon-delta definition effectively, which required:

• Understanding that as \( x \to \infty \), the term \( \sqrt{4x^2+1} \) is approximately equal to \( 2x \).
• Simplifying the function to a formula that can be handled analytically.

We typically tackle calculus problems like these by breaking down the function into simpler terms or approximations, which allows for the application of theorems like the epsilon-delta definition to find specific parameters efficiently.

Formal Definition of Limit

The formal definition of a limit underpins many concepts in calculus. When discussing limits at infinity, as in this problem, understanding this definition is foundational. The formal definition, or the epsilon-delta definition, states that for \( \lim_{x \to \infty} f(x) = L \), no matter how small \( \varepsilon \) is chosen, there is always an \( N \) after which all \( x > N \) will result in \( |f(x) - L| < \varepsilon \).

• This principle helps enable precise calculations and predictions.
• It's a bedrock of calculus, illustrating how functions behave over different domains.

Diving into the step-by-step process for the epsilon-delta approach sharpens comprehension of how even the smallest \( \varepsilon \) adjustments can dictate the outcome, demonstrating practical applications in finding accurate limits in real scenarios.