Problem 18

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. $$ f(x)=\sqrt{x}, \quad L=1 / 2, \quad c=1 / 4, \quad \epsilon=0.1 $$

Step-by-Step Solution

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Answer

The open interval is $(0.16, 0.36)$, and $\delta = 0.09$.

1Step 1: Understand the Limit Condition

The problem wants us to find an interval around $c = \frac{1}{4}$ such that $|f(x) - L| < \epsilon$ holds true, where $f(x) = \sqrt{x}$, $L = \frac{1}{2}$, and $\epsilon = 0.1$. Then, we have to find a $\delta$ such that for $0 < |x - c| < \delta$, $|f(x) - L| < \epsilon$ is satisfied.

2Step 2: Express the Condition as an Inequality

The inequality $|f(x) - L| < \epsilon$ becomes $|\sqrt{x} - \frac{1}{2}| < 0.1$. This inequality needs to hold true for the values of $x$ in terms of distance from $c = \frac{1}{4}$.

3Step 3: Solve the Inequality

To solve $|\sqrt{x} - \frac{1}{2}| < 0.1$, break it into two inequalities: 1. $\sqrt{x} - \frac{1}{2} < 0.1$ 2. $\sqrt{x} - \frac{1}{2} > -0.1$. These become:$1. \sqrt{x} < 0.6$ and $2. \sqrt{x} > 0.4$.Squaring both sides gives us:$0.16 < x < 0.36$. This is the interval where $|f(x) - L| < \epsilon$ is satisfied.

4Step 4: Determine $\delta$ Based on the Interval

We want to find $\delta$ such that $0 < |x-c| < \delta$ implies $0.16 < x < 0.36$. The center $c = \frac{1}{4} = 0.25$. Calculate the distance from 0.25 to each endpoint of the interval [0.16, 0.36]:- Distance from 0.25 to 0.16 is 0.09.- Distance from 0.25 to 0.36 is 0.11.So, $\delta = 0.09$, as it is the smaller value and ensures $|x-c| < \delta$ keeps $x$ within the interval.

Key Concepts

Epsilon-Delta DefinitionInequality SolvingInterval AnalysisFunction Limits

Epsilon-Delta Definition

The epsilon-delta definition is a formalization of the limits in mathematical analysis. It provides a precise way to show how a function approaches a certain value, known as the limit, as the input approaches a specific point.
Understanding the epsilon ($ \epsilon $) and delta ($ \delta $) components is crucial:

**Epsilon ($ \epsilon $)** represents a small positive number, indicating how close the function's value should be to the limit, $ L $.
**Delta ($ \delta $)** is another small positive number ensuring that the function's values are within an epsilon distance of the limit whenever the input is within a delta distance from the point, $ c $.

For example, we want $|f(x) - L| < \epsilon $to hold whenever $0 < |x-c| < \delta $. This means within this small $\delta $-distance from $c$, $f(x)$ remains very close to $L$.

Inequality Solving

Inequality solving is a method used in finding the specific values or intervals of a variable that satisfy an inequality condition.
This is critical when applying the epsilon-delta definition as we often have to solve inequalities of the form $|f(x) - L| < \epsilon$. You usually split the absolute value inequality $|\sqrt{x} - \frac{1}{2}| < 0.1 $into two components:

**Positive Component**: $\sqrt{x} - \frac{1}{2} < 0.1$
**Negative Component**: $\sqrt{x} - \frac{1}{2} > -0.1$

By solving each of these inequalities independently, you determine a range of values for $x$ which satisfy both conditions simultaneously. Solving these would suggest using operations such as taking roots or manipulating expressions to isolate $x$ within the resulting inequality.

Interval Analysis

Interval analysis refers to identifying specific regions in the domain of the function where a particular condition holds true. Here, the condition is $|f(x) - L| < \epsilon$.Interval analysis consists of finding such values of $x$, around $c$, which satisfy this criteria.

Upon solving the inequality $|\sqrt{x} - \frac{1}{2}| < 0.1$, we derived that the suitable interval for $x$ is within $0.16 < x < 0.36$.

This interval displays where the output of $f(x)$, specifically $\sqrt{x}$, remains sufficiently close to $L = \frac{1}{2}$.
The interval length relative to $c$ factors into determining $\delta$, making analysis crucial for further calculations.

Function Limits

Understanding the limits of a function is a foundational concept in calculus. A function's limit at some point $c$ is the value that the function approaches as $x$ gets arbitrarily close to $c$.In this exercise, the function $f(x) = \sqrt{x}$ approaches the limit $L = \frac{1}{2}$ as $x$ approaches $c = \frac{1}{4}$. Function limits establish predictability and understanding how functions behave near specific points:

**Evaluating Limits**: This includes finding where $f(x)$ approximates $L$.
**Continuity and Approximations**: Assessment of continuity helps in understanding whether $f(x)$ being close to $L$ translates to $x$ being close to $c$.

Using our example, the verification of $|\sqrt{x} - \frac{1}{2}| < 0.1$ within a certain range supports that $L$ is achieved when $x$ is near $c$.

Problem 18

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