Chapter 1: Problem 29
Use an \(\varepsilon-\delta\) proof to show that \(\lim _{x \rightarrow 1} 5 x-2=3\)
Short Answer
Expert verified
We use a \( \varepsilon-\delta \) proof, choosing \( \delta = \frac{\varepsilon}{5} \) to show the limit. The limit is 3.
Step by step solution
01
Write down the Definition of a Limit
To prove \( \lim_{x \to 1} (5x - 2) = 3 \) using an \( \varepsilon - \delta \) argument, start by recalling the definition of a limit: For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - 1| < \delta \), it follows that \( |(5x - 2) - 3| < \varepsilon \).
02
Simplify the Absolute Value Inequality
Simplify \( |5x - 2 - 3| < \varepsilon \).\[ |5x - 2 - 3| = |5x - 5| = 5|x - 1| \].
03
Relate x to \\varepsilon
For \( |5x - 5| < \varepsilon \), we want to express this in terms of \( |x - 1| \):\[ 5|x - 1| < \varepsilon \] implies \[ |x - 1| < \frac{\varepsilon}{5} \].
04
Choose \\delta
Choose \( \delta = \frac{\varepsilon}{5} \). This ensures that whenever \( |x - 1| < \delta \), \(|5x - 5| < \varepsilon\).
05
Formal Proof
For every \( \varepsilon > 0 \), let \( \delta = \frac{\varepsilon}{5} \). Then, if \( 0 < |x - 1| < \delta \), it follows that:\[ 5|x - 1| < 5\left(\frac{\varepsilon}{5}\right) = \varepsilon \].This confirms \( |(5x - 2) - 3| < \varepsilon \). Therefore, \( \lim_{x \to 1} (5x - 2) = 3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Limits
In calculus, a *limit* describes the behavior of a function as the input approaches a certain value. More specifically, it predicts the value that a function approaches as the input gets infinitely close to a particular point. This concept lays the foundation for much of calculus, helping to define derivatives and integrals.
- Limits help us understand the trends and behavior of functions.
- They allow us to handle expressions that may seem indeterminate at a specific point, often helping to simplify them.
Epsilon-Delta Definition of Limit
The *epsilon-delta definition* of a limit is a precise way to verify the value a function approaches as its input nears a particular point. This definition is fundamental in calculus because it provides a rigorous mathematical framework for limits.
For a function \( f(x) \), the statement \( \lim_{x \to c} f(x) = L \) means that for any tiny positive number \( \varepsilon \) (epsilon), no matter how small, there is a corresponding \( \delta \) (delta) such that whenever the distance between \( x \) and \( c \) is less than \( \delta \) (but not zero), the distance between \( f(x) \) and \( L \) is less than \( \varepsilon \).
For a function \( f(x) \), the statement \( \lim_{x \to c} f(x) = L \) means that for any tiny positive number \( \varepsilon \) (epsilon), no matter how small, there is a corresponding \( \delta \) (delta) such that whenever the distance between \( x \) and \( c \) is less than \( \delta \) (but not zero), the distance between \( f(x) \) and \( L \) is less than \( \varepsilon \).
- \( \varepsilon \) represents how close the output (\( f(x) \)) should be to the limit \( L \).
- \( \delta \) represents how close \( x \) should be to \( c \) to achieve this closeness in \( f(x) \).
Absolute Value Inequality in Epsilon-Delta Proofs
*Absolute value inequality* is crucial in epsilon-delta proofs because it quantifies how close two numbers are regardless of direction. In this exercise, understanding how to manipulate the absolute value helps clarify the relationship between \( x \) and the function's limit.
We want \( |(5x - 2) - 3| < \varepsilon \). Simplification leads to \( |5x - 5| = 5|x - 1| \). Thus, if \( 5|x - 1| < \varepsilon \), it means \( |x - 1| < \frac{\varepsilon}{5} \).
We want \( |(5x - 2) - 3| < \varepsilon \). Simplification leads to \( |5x - 5| = 5|x - 1| \). Thus, if \( 5|x - 1| < \varepsilon \), it means \( |x - 1| < \frac{\varepsilon}{5} \).
- This step ensures that the deviation of \( 5x - 2 \) from 3 is smaller than \( \varepsilon \).
- The manipulation of the absolute value symbol is vital to arrive at the appropriate \( \delta \).
Calculus and Epsilon-Delta Proofs
*Calculus* explores the concepts of limits, derivatives, integrals, and infinite series. The epsilon-delta proof is a quintessential example of calculus rigor, providing a thorough verification process for limits. This proof method ensures that the calculated limit holds true in all infinitesimally small evaluations.
In this exercise, the epsilon-delta proof steps through the logical reasoning needed in calculus:
In this exercise, the epsilon-delta proof steps through the logical reasoning needed in calculus:
- Determining \( \varepsilon \) sets the tolerance for output variation.
- Calculating \( \delta \) establishes the input range necessary to keep function values within this tolerance.