Problem 65 - Calculus Early Transcendentals

Question

$$ \begin{array}{c}{\text { (a) How large do we have to take } x \text { so that } 1 / x^{2}<0.0001 ?} \\ {\text { (b) Taking } r=2 \text { in Theorem } 5, \text { we have the statement }} \\ {\quad \lim _{x \rightarrow \infty} \frac{1}{x^{2}}=0} \\ {\text { Prove this directly using Definition } 7}\end{array} $$

Step-by-Step Solution

Answer

(a) \( x > 100 \). (b) Use \( N = \sqrt{\frac{1}{\epsilon}} \) for \( \epsilon > 0 \).

Step 1: Understand the inequality

To solve part (a), you need to find the value of \( x \) such that \( \frac{1}{x^2} < 0.0001 \). This means we need to rearrange the inequality to solve for \( x \).

Step 2: Rearrange the inequality

Starting with \( \frac{1}{x^2} < 0.0001 \), multiply both sides of the inequality by \( x^2 \) to get: \( 1 < 0.0001 \times x^2 \).

Step 3: Solve for \( x^2 \)

Divide both sides by 0.0001: \( x^2 > \frac{1}{0.0001} = 10000 \).

Step 4: Solve for \( x \)

Take the square root of both sides: \( x > \sqrt{10000} \). The square root of 10000 is 100, so \( x > 100 \).

Step 5: Prove the limit using the epsilon-delta definition

For part (b), we need to show that \( \lim_{{x \to \infty}} \frac{1}{x^2} = 0 \) using the precise \( \epsilon \)-\( N \) definition of a limit. We want to show for every \( \epsilon > 0 \), there exists an \( N \) such that for all \( x > N \), the inequality \( \left| \frac{1}{x^2} - 0 \right| < \epsilon \) holds.

Step 6: Set up the limit definition

We start with \( \left| \frac{1}{x^2} \right| < \epsilon \). This boils down to \( \frac{1}{x^2} < \epsilon \).

Step 7: Solve for \( x \) in terms of \( \epsilon \)

Rearrange the inequality to solve for \( x^2 \): \( x^2 > \frac{1}{\epsilon} \).

Step 8: Define N in terms of epsilon

To satisfy the limit condition, choose \( N = \sqrt{\frac{1}{\epsilon}} \). Then, for all \( x > N \), \( x^2 > \frac{1}{\epsilon} \), ensuring \( \left| \frac{1}{x^2} \right| < \epsilon \).

Key Concepts

epsilon-delta definition

In calculus, understanding the convergence of a function is crucial. The epsilon-delta definition is a formal way to show how functions behave as they approach a limit. When we say that the limit of a function \( f(x) \) as \( x \) approaches infinity is \( L \), the epsilon-delta definition provides a clear criterion for this behavior. Specifically, for a given \( \epsilon > 0 \), there should be a corresponding \( N \) such that for all \( x > N \), the inequality \( |f(x) - L| < \epsilon \) holds true. This framework helps us ensure the function's values get arbitrarily close to \( L \) as \( x \) increases without bound.

To apply this to infinite limits, like \( \lim_{{x \to \infty}} \frac{1}{x^2} = 0 \), it means for any positive \( \epsilon \), we can find an \( N \) after which the function's value is smaller than \( \epsilon \). This logical structure fosters a robust understanding of limits and guarantees precision in calculus proofs.

inequality solving

Inequality solving is a fundamental skill in calculus and mathematics generally. It involves determining the range of values that satisfy a given inequality. In the original exercise, to find \( x \) such that \( \frac{1}{x^2} < 0.0001 \), you must rearrange and solve the inequality.

Start by multiplying both sides by \( x^2 \) to remove the fraction: \( 1 < 0.0001x^2 \). Next, isolate \( x^2 \) by dividing both sides by 0.0001: \( x^2 > 10000 \). Finally, solve for \( x \) by taking the square root, yielding \( x > 100 \).

This approach to solving inequalities is a powerful tool. It can help reveal conditions under which specific mathematical statements hold true. By learning to manipulate and solve inequalities, students can tackle a wide variety of mathematical problems efficiently.

infinite limits

Infinite limits describe the behavior of functions as the variable tends towards positive or negative infinity. When dealing with limits like \( \lim_{{x \to \infty}} \frac{1}{x^2} = 0 \), we're interested in what happens to \( \frac{1}{x^2} \) as \( x \) becomes very large.

As \( x \) increases, \( \frac{1}{x^2} \) becomes smaller and smaller, approaching 0. The notion of an infinite limit doesn't imply that the function reaches 0, but rather it gets closer without bound. This concept is vital in understanding various real-world processes described by mathematical models, including decay rates and asymptotic analysis.

limit proofs

Proving limits can often seem daunting, but with practice, it becomes manageable. Limit proofs, especially using the epsilon-delta definition, require a logical sequence of reasoning. The goal is to show that for every \( \epsilon > 0 \), there's a point beyond which the distance between \( f(x) \) and the limit \( L \) stays within \( \epsilon \).

Let's consider the limit \( \lim_{{x \to \infty}} \frac{1}{x^2} = 0 \). To prove this, start by stating \( |\frac{1}{x^2} - 0| < \epsilon \). This simplifies to \( \frac{1}{x^2} < \epsilon \). From here, reverse engineer the inequality to find \( x^2 > \frac{1}{\epsilon} \), leading to \( x > \sqrt{\frac{1}{\epsilon}} \).

Define \( N = \sqrt{\frac{1}{\epsilon}} \). Therefore, for all \( x > N \), \( \frac{1}{x^2} < \epsilon \). This method transforms the abstract notion of limits into a concrete framework, allowing students to craft rigorous mathematical proofs.