Problem 21 - Thomas Calculus

Question

Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$ f(x)=1 / x, \quad L=1 / 4, \quad c=4, \quad \epsilon=0.05 $$

Step-by-Step Solution

Answer

Choose interval \(\left(\frac{10}{3}, 5\right)\), let \(\delta = 0.25\).

Step 1: Understand the Function and Limits

We have the function \( f(x) = \frac{1}{x} \) and need to find an interval around \( c = 4 \) such that \( |f(x) - L| < \epsilon \), where \( L = \frac{1}{4} \) and \( \epsilon = 0.05 \).

Step 2: Set Up the Inequality

Start with the inequality: \[ |f(x) - L| < \epsilon \] Substitute the given values: \[ \left| \frac{1}{x} - \frac{1}{4} \right| < 0.05 \].

Step 3: Solve the Inequality for \(x\)

Reformulate the inequality: \[ \left| \frac{4 - x}{4x} \right| < 0.05 \]To solve, find when \( -0.05 < \frac{4-x}{4x} < 0.05 \). Solve these two inequalities separately.

Step 4: Solve \(-0.05 < \frac{4-x}{4x}\)

Multiply through by \(4x\), assuming \(x eq 0\), to get:\[ -0.05 \cdot 4x < 4 - x \] \[ -0.2x < 4 - x \] \[ 0.8x > 4 \] Thus, \( x > 5 \). However, this result is incorrect for our purposes since it moves in wrong direction.

Step 5: Solve \(\frac{4-x}{4x} < 0.05\)

Using the second part:\[ \frac{4-x}{4x} < 0.05 \] Multiply through by \(4x\):\[ 4 - x < 0.2x \] \[ 4 < 1.2x \] \[ x > \frac{10}{3} \].

Step 6: Combine Results and Find Interval

Combining solutions gives \( \frac{10}{3} < x < 5 \). However, \(5\) is unreasonable since we must be near \(4\), so adjust considering opening around \(4\).

Step 7: Choosing \(\delta\)

To find \( \delta \), we have established \( \frac{10}{3} < x < 5 \). We choose a small \( \delta = 0.25 \) so that\(0 < |x - 4| < \delta \) is satisfied by this interval.

Key Concepts

Epsilon-Delta Definition

Understanding the epsilon-delta definition of a limit is crucial to grasping the behavior of functions as they approach a certain value. In the context of limits, \( \varepsilon \-delta\) definition is a formal mathematical way to express how close we want a function to be to a certain limit. We use \( \varepsilon \) and \( \delta \) to achieve a precise description:

• \( \varepsilon \) (epsilon) can be thought of as a small distance around the limiting value \( L \).
• \( \delta \) (delta) represents a small interval around the point \( c \) such that all values of the function within this interval are close to \( L \).

For example, for the function \( f(x) = \frac{1}{x} \) as \( x \) approaches \( 4 \), we ensure that for every small \( \varepsilon \), there exists a \( \delta \) such that within the interval \( 0 < |x - 4| < \delta \), the function \( f(x) \) remains within \( \varepsilon \) distance from the limit \( L = \frac{1}{4} \). By finding such intervals and \( \delta \) values, we ensure continuity and better understand functional behavior.

Continuity of Functions

A function is continuous at a point \( c \) if it smoothly flows through without jumping or having undefined points at that location. More formally, a function \( f(x) \) is continuous at \( x = c \) if all three conditions are met:

• The function \( f(x) \) is defined at \( x = c \).
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

In our exercise, we analyze the function \( f(x) = \frac{1}{x} \) near the point \( c = 4 \). We ensure that when \( x \) is near 4, the function value remains close to \( L = \frac{1}{4} \), making it practically smooth and uninterrupted around this point. This concept of continuity ensures there are no sudden jumps or discontinuities in the function's behavior in this interval. Whenever \( x \) is within a delta range of 4, the function behaves as expected, gliding smoothly without interruption.

Solving Inequalities

Solving inequalities is about finding the set of all values of \( x \) that make the inequality true. An inequality such as \( |f(x) - L| < \varepsilon \) requires us to determine conditions on \( x \) that keep the function's value near \( L \).In our exercise, we transformed the inequality \( \left| \frac{1}{x} - \frac{1}{4} \right| < 0.05 \) into a more manageable form. We first simplified the problem to tackle both sides:

• We dealt with \( \frac{4-x}{4x} < 0.05 \), and \( -0.05 < \frac{4-x}{4x} \).

By splitting and solving these inequalities, we determined the range for \( x \) close to \( c = 4 \) that satisfies the conditions set by \( \varepsilon \). Eventually, combining these, we found the interval \( \frac{10}{3} < x < 5 \). However, considering practical aspects, the interval tighter around 4 is what ensures our requirements, demonstrating how inequalities guide us to find appropriate intervals where the function resides within a desired proximity to its limit.