Problem 16 - The Calculus with Analytic Geometry

Question

Prove that \(\lim _{x \rightarrow+\infty} f(x)=1\) by applying Definition 4.1.1; that is, for any \(\epsilon>0\), show that there exists a number \(N>0\) such that \(|f(x)-1|<\epsilon\) whenever \(x>N\). $$ f(x)=\frac{x^{2}+2 x}{x^{2}-1} $$

Step-by-Step Solution

Answer

For any \epsilon > 0, choose N=2/\epsilon. Then |f(x)-1| < \epsilon whenever x > N.Therefore, lim _{x \rightarrow+\infty} f(x)=1.

Step 1: Restate the Limit Definition

To prove that \( lim _{x \rightarrow+\infty} f(x)=1 \) using Definition 4.1.1, we need to show that for any \( \epsilon>0 \), there exists a number \( N>0 \) such that \( |f(x)-1|<\epsilon \) whenever \( x>N \).

Step 2: Simplify \(f(x)\)

Let's simplify \( f(x)\): \[ f(x)=\frac{x^{2}+2 x}{x^{2}-1} \] This can be rewritten as: \[ f(x)= \frac{x^{2}+2x}{x^{2}-1} = \frac{x^{2}+2x}{x^{2}(1-\frac{1}{x^{2}})} = \frac{x^{2}+2x}{x^{2}} * \frac{1}{1-\frac{1}{x^{2}}} \]

Step 3: Simplify Further

Continuing from where we left off: \[ f(x)= (1 + \frac{2}{x}) * \frac{1}{1-\frac{1}{x^{2}}} \] As \( x \rightarrow +\infty \), \( \frac{2}{x} \) approaches 0 and \( \frac{1}{x^{2}} \) approaches 0. Thus, \[ f(x) \rightarrow (1+0) * \frac{1}{1-0} = 1 \]

Step 4: Establish \(|f(x) - 1| < \epsilon\)

Given \( \epsilon > 0 \), we need to establish that \( |f(x)-1|<\epsilon \) for \( x>N \). We have: \[ | (1 + \frac{2}{x} - 1) * \frac{1}{ 1- \frac{1}{x^{2}}} | = | \frac{2}{x} * \frac{1}{ 1- \frac{1}{x^{2}}} | = \frac{2}{x( 1- \frac{1}{x^{2}} )} \] We know that as \( x \rightarrow + \infty \) tends \( \frac{1}{x^{2}} \) gets smaller approaching 0, hence \( 1 - \frac{1}{x^{2}} \rightarrow 1 \) and the fraction \( \frac{2}{x( 1- \frac{1}{x^{2}} )} \rightarrow \frac{2}{x} \) and thus given \( x>N= \frac{2}{\epsilon}\) \( | \frac{2}{x( 1- \frac{1}{x^{2}} )} | < \epsilon \) is true if and only if, x>N.

Step 5: Conclusion

Therefore, we have proven that for any \( \epsilon > 0 \), there exists an \( N > 0 \), specifically \( N= \frac{2}{\epsilon} \), such that if \( x > N \), \( |f(x) - 1| < \epsilon \). Hence, \( \lim_{x \rightarrow +\infty} f(x) = 1 \) is true.

Key Concepts

Epsilon-Delta Definition

To understand the epsilon-delta definition, first grasp the basic idea behind limits. Limits help us understand the behavior of a function as its input approaches a certain value. The epsilon-delta definition provides a precise way to express this idea.
In simple terms, for the limit of a function \( f(x) \) to be 1 as \( x \) approaches infinity, given any small positive number \( \epsilon \), no matter how small, there exists a large number \( N \) such that whenever \( x \) is larger than \( N \), the value of \( f(x) \) will be within an \( \epsilon \) distance from 1.
Breaking it down:

• \( \epsilon \) (epsilon) is any positive small number you choose.
• \( N \) is a large number dependent on \( \epsilon \).
• The function value \( f(x) \) needs to be within the interval (1 - \( \epsilon \), 1 + \( \epsilon \)).

For our function \( f(x) = \frac{x^{2}+2x}{x^{2}-1} \), we showed that for given \( \epsilon \), choosing \( N = \frac{2}{\epsilon} \) ensures that \( |f(x) - 1| < \epsilon \) when \( x > N \). This completes the epsilon-delta proof.

Limits at Infinity

When dealing with limits at infinity, we look at the behavior of a function as its input grows without bound. The key idea is to determine where the function stabilizes as \( x \) becomes very large.
For the given function \( f(x) = \frac{x^{2}+2x}{x^{2}-1} \), we are interested in its behavior as \( x \) goes to positive infinity. Simplifying the function:
We reframe it as:
\[ f(x) = \frac{x^{2}+2x}{x^{2}-1} = \frac{x^{2}+2x}{x^{2}(1 - \frac{1}{x^{2}})} = (1 + \frac{2}{x}) * \frac{1}{1 - \frac{1}{x^{2}}} \] Notice how as \( x \) grows larger, the terms \( \frac{2}{x} \) and \( \frac{1}{x^{2}} \) approach zero. This simplifies further to \[ f(x) \approx (1+0) \cdot \frac{1}{1-0} = 1 \].
Therefore, as \( x \) approaches infinity, \( f(x) \) approaches 1. Hence, \( \lim_{x \to \infty} f(x) = 1 \) is confirmed by analysis of its algebraic simplification.

Calculus and its Application

Calculus is essential for understanding rates of change and gradients. It forms a core part of the mathematical toolkit for analyzing functions and their behaviors.
In this exercise, we focused on one such application: proving a limit using rigorous methods. Specifically, we used the epsilon-delta definition to verify the behavior of \( f(x) \) at infinity.
Calculus allows us to:

• Calculate precise derivatives and integrals.
• Understand continuity and differentiability of functions.
• Apply limits to solve real-world problems, such as optimizing functions.

By mastering these concepts, students gain a deeper appreciation of the mathematical language used to describe natural phenomena and solve complex problems. This exercise illustrates a practical application of calculus concepts, reinforcing their understanding of limits and function behavior. It’s a foundational skill for delving into more advanced topics in mathematics and related fields.