Problem 28 - 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. $$ f(x)=m x, \quad m>0, \quad L=3 m, \quad c=3, \quad \epsilon=c>0 $$

Step-by-Step Solution

Answer

Interval: \((3 - \frac{3}{m}, 3 + \frac{3}{m})\); \(\delta = \frac{3}{m}\).

Step 1: Identify the Components

We have the function \( f(x) = mx \), where \( m > 0 \). The given values are \( L = 3m \), \( c = 3 \), and \( \epsilon = 3 \). Our goal is to find an interval around \( c = 3 \) where the condition \( |f(x) - L| < \epsilon \) is satisfied.

Step 2: Set Up the Inequality

Start with the inequality \(|f(x) - L| < \epsilon\). Substitute the appropriate expressions: \(|mx - 3m| < 3\). This can be simplified to \(m|x - 3| < 3\). By dividing both sides by \(m\) (since \(m > 0\)), we get the inequality \(|x - 3| < \frac{3}{m}\).

Step 3: Define the Open Interval

From \(|x - 3| < \frac{3}{m}\), the interval is determined. Therefore, the open interval about \( c \) is \( (3 - \frac{3}{m}, 3 + \frac{3}{m}) \). Within this interval, \(|f(x) - L| < \epsilon\) holds.

Step 4: Determine a Suitable Delta

To satisfy \( 0 < |x - 3| < \delta \) and \(|f(x) - L| < \epsilon\), set \( \delta = \frac{3}{m} \). Thus, for all \( x \) such that \( 0 < |x - 3| < \delta \), the inequality \(|mx - 3m| < 3\) is satisfied.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way of defining the limit of a function. It involves two key quantities: \( \epsilon \) and \( \delta \).

• \( \epsilon \): This is a positive number, often small, representing how close the value of a function \( f(x) \) should be to the limit \( L \).
• \( \delta \): This is another positive number, defining the range around the point \( c \) within which all \( x \) should satisfy the condition for \( \epsilon \).

The definition states that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x-c| < \delta \), then \( |f(x) - L| < \epsilon \). The goal is to "trap" \( f(x) \) close to \( L \) by restricting \( x \) close to \( c \). This method ensures we can find a small interval around \( c \) where the function behaves predictably, confirming its limit at that point.

Open Interval

An open interval is a crucial concept in understanding limits and continuity. In the context of limits, an open interval around a number \( c \) is the set of points that are close to \( c \) but not including \( c \) itself.

• Representation: An open interval surrounding \( c \) is expressed as \((c - a, c + a)\), where \( a > 0 \).
• Properties: The boundary points \( c - a \) and \( c + a \) are not part of the interval.

In this exercise, examining the open interval \((3 - \frac{3}{m}, 3 + \frac{3}{m})\) helps us ensure that the function values remain within an acceptable range of \( L = 3m \). This surrounds \( c = 3 \) and makes sure the difference \(|f(x) - L|\) stays less than \( \epsilon \), fulfilling the condition for continuity.

Function Continuity

Function continuity is another important aspect when considering limits. A function is said to be continuous at a point \( c \) if, loosely speaking, you can draw its graph without lifting your pencil at \( c \).Here's how continuity relates to the epsilon-delta definition:

• Purpose: Demonstrates that small changes in \( x \) near \( c \) produce small changes in \( f(x) \).
• Criteria: A function \( f(x) \) is continuous at \( x = c \) if \( \lim_{{x \to c}} f(x) = f(c) \).

In our problem, by identifying an appropriate \( \delta \), we ensure that \( f(x) = mx \) behaves continuously around \( x = 3 \). As \( x \) approaches 3 (but not exactly 3), the output \( mx \) remains stably within a bounded distance \( \epsilon = 3 \) of \( L = 3m \), confirming the function's continuity at that point.