Problem 14
Question
Given that \( \displaystyle \lim_{x \to 2}(5x - 7) = 3 \), illustrate Definition 2 by finding values of \( \delta \) that correspond to \( \varepsilon = 0.1 \), \( \varepsilon = 0.05 \), and \( \varepsilon = 0.01 \).
Step-by-Step Solution
Verified Answer
For \( \varepsilon = 0.1 \), \( \delta = 0.02 \); for \( \varepsilon = 0.05 \), \( \delta = 0.01 \); for \( \varepsilon = 0.01 \), \( \delta = 0.002 \).
1Step 1: Understand the Definition of Limit
The definition states that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \), then \(|(5x - 7) - 3| < \varepsilon \). We need to find corresponding \( \delta \) values for given \( \varepsilon \) values.
2Step 2: Rewrite the Limit Condition
Start with \( |(5x - 7) - 3| < \varepsilon \). Simplify it to \( |5x - 10| < \varepsilon \). This implies \( 5|x - 2| < \varepsilon \).
3Step 3: Solve for \( |x - 2| \)
Divide both sides of \( 5|x - 2| < \varepsilon \) by 5 to find \( |x - 2| < \frac{\varepsilon}{5} \). The expression \( \frac{\varepsilon}{5} \) gives the corresponding \( \delta \) value for a specific \( \varepsilon \).
4Step 4: Solve for Specific \( \varepsilon \) Values
Now apply the formula from Step 3 to each \( \varepsilon \):- If \( \varepsilon = 0.1 \), then \( \delta = \frac{0.1}{5} = 0.02 \).- If \( \varepsilon = 0.05 \), then \( \delta = \frac{0.05}{5} = 0.01 \).- If \( \varepsilon = 0.01 \), then \( \delta = \frac{0.01}{5} = 0.002 \).
Key Concepts
Limit ProofEpsilon-Delta CriterionCalculus Limits
Limit Proof
Understanding how to prove a limit is essential in calculus. A limit proof shows that as a variable approaches a specific value, a function gets closer to a particular number. It's like proving that as you walk toward a destination, you're getting closer to it step by step. The epsilon-delta proof is a standard method used in this process.
Proving a limit involves several steps:
Proving a limit involves several steps:
- Identify the function and the point as the variable approaches.
- Formulate what it means to get 'close enough.' This involves the quantities epsilon (\(\varepsilon\)) and delta (\(\delta\)).
- Show that for every \(\varepsilon > 0\), we can find a \(\delta > 0\), the condition \( 0 < |x - 2| < \delta \) results in \(|f(x) - L| < \varepsilon\), where \(L\) is the limit value.
- Work through the specific example, determining which \(\delta\) values fit the given \(\varepsilon\) conditions.
Epsilon-Delta Criterion
The epsilon-delta criterion is at the heart of understanding the formal definition of a limit. This method might seem complex initially, but it's a powerful way to confirm that a function approaches a limit. Here's a simpler breakdown:
- **Epsilon (\(\varepsilon\))** represents how close the function's output (or \( f(x) \)) needs to be to the limit value.
- **Delta (\(\delta\))** corresponds to how near the input \(x\) must be to the target value to meet the epsilon condition.
To utilize the epsilon-delta criterion:
- **Epsilon (\(\varepsilon\))** represents how close the function's output (or \( f(x) \)) needs to be to the limit value.
- **Delta (\(\delta\))** corresponds to how near the input \(x\) must be to the target value to meet the epsilon condition.
To utilize the epsilon-delta criterion:
- Begin by expressing the desired closeness of the function to the limit, given by \(|f(x) - L| < \varepsilon\).
- Determine a distance \(\delta\) for \(x\) so that \(0 < |x - c| < \delta\) enforces the inequality above.
Calculus Limits
Limits form the foundational pillar in calculus, essential for understanding concepts like continuity, derivatives, and integrals. By grasping limits, students unlock a deeper comprehension of calculus as a whole. A limit essentially tells us what happens to a function as inputs get closer and closer to a specified point.
There are several types of limits:
There are several types of limits:
- **Finite Limits**: When a function approaches a specific real number as \(x\) approaches a certain point.
- **Infinite Limits**: Occur when a function grows without bounds as \(x\) approaches a particular value.
- **Limits at Infinity**: Describe the behavior of a function as \(x\) increases or decreases without bound.
Other exercises in this chapter
Problem 14
Evaluate the limit and justify each step by indicating the appropriate properties of limits. \( \displaystyle \lim_{x \to \infty} \sqrt{\dfrac{9x^3 + 8x - 4}{3
View solution Problem 14
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number \( a \). \( f(x) = 3x^4 - 5x + \sqrt[3
View solution Problem 14
Evaluate the limit, if it exists. \( \displaystyle \lim_{x \to 4}\frac{x^2 + 3x}{x^2 - x - 12} \)
View solution Problem 15
The displacement (in meters) of a particle moving in a straight line is given by the equation of motion \( s = 1/t^2 \), where \( t \) is measured in seconds. F
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