Problem 24 - 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. $$ f(x)=1 / x, \quad L=-1, \quad c=-1, \quad \epsilon=0.1 $$

Step-by-Step Solution

Answer

Interval: \((-1.11, -0.91)\), \(\delta = 0.1\).

Step 1: Analyze the Given Function and Limits

We have the function \( f(x) = \frac{1}{x} \), given limit \( L = -1 \), and center \( c = -1 \). The task is to identify an interval around \( c \) where \(|f(x) - L| < \epsilon \), with \( \epsilon = 0.1 \).

Step 2: Set Up the Inequality

We want the condition \(|f(x) - L| < \epsilon\) to hold. Substituting in the known values, this becomes \(|\frac{1}{x} + 1| < 0.1\).

Step 3: Solve the Inequality \(|\frac{1}{x} + 1| < 0.1\)

Rewriting and breaking down the inequality, we have two conditions: 1. \(\frac{1}{x} + 1 < 0.1\) 2. \(-0.1 < \frac{1}{x} + 1\) For \(\frac{1}{x} + 1 < 0.1\):\[ \frac{1}{x} < -0.9 \]\[ x < -\frac{1}{0.9} = -\frac{10}{9} \approx -1.11 \]For \(-0.1 < \frac{1}{x} + 1\):\[ -1.1 < \frac{1}{x} \]\[ x > -\frac{1}{1.1} = -\frac{10}{11} \approx -0.91 \]

Step 4: Determine the Open Interval Around \(c\)

From the inequality solutions, the open interval around \(c = -1\) satisfying \(|f(x) - L| < \epsilon\) is approximately \((-\frac{10}{9}, -\frac{10}{11})\), which can be simplified to \((-1.11, -0.91)\).

Step 5: Choose and Verify \(\delta\)

The interval \((-1.11, -0.91)\) is symmetric around \(c = -1\) with width \(0.1\) on each side. Therefore, we can choose \(\delta = 0.1\). To verify, ensure for any \(x\) in \((-1 - \delta, -1 + \delta)\), \(|\frac{1}{x} - (-1)| < 0.1\) remains true within the interval.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a rigorous way to define the limit of a function. It articulates the idea that for a function \( f(x) \) to approach a limit \( L \) as \( x \) approaches \( c \), the difference between \( f(x) \) and \( L \) can be made arbitrarily small by choosing \( x \) sufficiently close to \( c \). This can be expressed as:

• For every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \epsilon \).

Here, \( \epsilon \) represents how close we want \( f(x) \) to be to \( L \), and \( \delta \) tells us how close \( x \) should be to \( c \) to achieve this closeness.
By solving \( |f(x) - L| < \epsilon \), we find an appropriate \( \delta \) that guarantees \( f(x) \) remains within \( \epsilon \) distance from \( L \). In our example, solving \( |\frac{1}{x} + 1| < 0.1 \) provided us a \( \delta \) such that this inequality holds true around \( c = -1 \).

Continuity

Continuity is an essential concept that ties into the behavior of functions at specific points. A function is continuous at point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) equals \( f(c) \). In terms of epsilon-delta definition, this means:

• For every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - f(c)| < \epsilon \).

This assures that a graph of a continuous function will have no breaks, holes, or jumps at \( x = c \).
In our initial task, the point \( c = -1 \) serves as a focus to determine the interval in which our function maintains proximity to the limit \( L = -1 \), indicating continuous-like behavior in that restricted domain despite that \( \frac{1}{x} \) is not continuous at \( c \) due to its undefined nature at 0.

Open Intervals

Open intervals are crucial when analyzing limits and continuity, as they define a range of \( x \) values that exclude endpoints. In mathematics, the notation \( (a, b) \) represents all numbers \( x \) where \( a < x < b \), but does not include \( a \) and \( b \) themselves.

• This exclusivity makes open intervals useful in limit problems where exact points might not be included or relevant to the function's behavior, such as when a function approaches but never actually reaches a limit value at its endpoints.

In the exercise, the interval \( (-1.11, -0.91) \) is an open interval surrounding \( c = -1 \).
This interval symbolizes where \( \frac{1}{x} \) can be close to \( L = -1 \) without including the endpoint values themselves, aligning with our goal using the epsilon-delta definition to stay within \( \epsilon = 0.1 \) from \( L \).
Open intervals help conceptualize how closely a function can approach a given limit in limit problems, while ensuring functions remain well-defined and logically consistent.