Chapter 2: Problem 8
For the limit $$ \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}=2 $$ illustrate Definition 2 by finding values of \(\delta\) that correspond to \(\varepsilon=0.5\) and \(\varepsilon=0.1\)
Short Answer
Expert verified
\(\delta = 0.5\) for \(\varepsilon=0.5\), \(\delta ≈ 0.2236\) for \(\varepsilon=0.1\).
Step by step solution
01
Understand the Problem Statement
The exercise asks us to use the definition of a limit to find values of \(\delta\) such that whenever \(0 < |x - 0| < \delta\), it guarantees that \(\left| \frac{e^{2x} - 1}{x} - 2 \right| < \varepsilon\) for given \(\varepsilon\) values. \(\varepsilon\) is a measure of how close we want \(\frac{e^{2x} - 1}{x}\) to be to 2. In this exercise, we're given specific \(\varepsilon\) values: 0.5 and 0.1.
02
Use Limit Definition to Set Up the Inequality
According to the limit definition, we know \(\lim _{x \rightarrow 0} \frac{e^{2x}-1}{x}=2\). This means for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x| < \delta\), then \(\left| \frac{e^{2x} - 1}{x} - 2 \right| < \varepsilon\). We start by setting up the inequality: \(\left| \frac{e^{2x} - 1}{x} - 2 \right| < \varepsilon\).
03
Simplify the Inequality Expression
Substitute and simplify the inequality: \(\left| \frac{e^{2x} - 1}{x} - 2 \right|= \left| \frac{e^{2x} - 1 - 2x}{x} \right| < \varepsilon\). Here, note that a Maclaurin expansion of \(e^{2x}\) can be used to approximate \(e^{2x} \approx 1 + 2x + 2x^2/2 + \cdots\) for small \(x\), which simplifies to \(e^{2x} - 1 ≈ 2x + 2x^2\). Substitute back: \(\left| 2x^2 / x \right| < \varepsilon\), this simplifies to \(2x < \varepsilon\).
04
Solve for \(\delta\) for \(\varepsilon=0.5\)
For \(\varepsilon = 0.5\): We have \(2x^2 < 0.5|x|\). Solving for \(|x|\), we have \(|x| < \sqrt{\varepsilon/2} = \sqrt{0.25} = 0.5\). Thus, \(\delta = 0.5\) when \(\varepsilon = 0.5\).
05
Solve for \(\delta\) for \(\varepsilon=0.1\)
For \(\varepsilon = 0.1\): Use the same process, \(2x^2 < 0.1|x|\). Solving for \(|x|\), we have \(|x| < \sqrt{\varepsilon/2} = \sqrt{0.05} ≈ 0.2236\). Therefore, \(\delta = 0.2236\) when \(\varepsilon = 0.1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
calculus limits
Calculus limits are fundamental in understanding how functions behave near specific points. A limit helps us comprehend what value a function approaches as the input gets infinitely close to a particular number.
The formal definition involves two values, \( ext{ε (epsilon)} \) and \( ext{δ (delta)} \), which serve to illustrate precision and proximity. Specifically, when we say the limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \), this implies that for every \( ext{ε} > 0\), however small, there exists a \( ext{δ} > 0\) such that \(|f(x) - L| < ext{ε}\) whenever \(0 < |x - c| < ext{δ}\).
The formal definition involves two values, \( ext{ε (epsilon)} \) and \( ext{δ (delta)} \), which serve to illustrate precision and proximity. Specifically, when we say the limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \), this implies that for every \( ext{ε} > 0\), however small, there exists a \( ext{δ} > 0\) such that \(|f(x) - L| < ext{ε}\) whenever \(0 < |x - c| < ext{δ}\).
- \( ext{Epsilon (ε)} \) defines how close \( f(x) \) has to be to \( L \).
- \( ext{Delta (δ)} \) indicates how close \( x \) should be to \( c \).
limit approximation
Limit approximation is often employed when an exact calculation is intricate or impossible. It involves techniques to estimate the limit of a function more easily. In the case of complex functions, such as exponential functions, approximation can simplify the process significantly.
For small values of \( x \), the exponential function \( e^{2x} \) is approximately \( 1 + 2x + 2x^2 \) using the Maclaurin series. This approximation aids in understanding the behavior of \( rac{e^{2x} - 1}{x} \) as \( x \) approaches zero.
For small values of \( x \), the exponential function \( e^{2x} \) is approximately \( 1 + 2x + 2x^2 \) using the Maclaurin series. This approximation aids in understanding the behavior of \( rac{e^{2x} - 1}{x} \) as \( x \) approaches zero.
- This approximation transforms complicated expressions into simpler, more manageable forms.
- By approximating, we achieve a practical insight into the limit, facilitating the calculation of corresponding \( δ \) values for given \( ε \) values.
Maclaurin series
The Maclaurin series is a special case of the Taylor series, specifically centered at zero. It provides a way to express functions as infinite sums of terms. For functions like \( e^{2x} \), the Maclaurin series is very helpful in limit approximation.
The series expansion for \( e^{2x} \) is:\[ e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots \]
For small values of \( x \), higher-degree terms become negligible, making the series:\[ e^{2x} \approx 1 + 2x + 2x^2 \]
The series expansion for \( e^{2x} \) is:\[ e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots \]
For small values of \( x \), higher-degree terms become negligible, making the series:\[ e^{2x} \approx 1 + 2x + 2x^2 \]
- This simplification helps solve limits efficiently without evaluating complicated exponential functions each time.
- The Maclaurin series provides clarity in polynomial approximation, letting us handle originally complex problems with ease.
calculus problem solving
Solving calculus problems, especially those involving limits, often requires a strategic approach. The epsilon-delta definition provides a rigorous method, yet the solution process can be streamlined with practical techniques like approximations and series.
- Step-by-step evaluations, such as setting inequalities based on the limit definition, bring clarity.
- Approximation using series, like the Maclaurin expansion, helps in simplifying complex expressions into polynomial forms for easy evaluation.
- Understanding how to choose appropriate \( ext{δ} \) values for given \( ext{ε} \) choices underlies problem-solving efficiency.