Problem 29 - APEX Calculus

Question

Use an \(\varepsilon-\delta\) proof to show that \(\lim _{x \rightarrow 1} 5 x-2=3\)

Step-by-Step Solution

Answer

We use a \( \varepsilon-\delta \) proof, choosing \( \delta = \frac{\varepsilon}{5} \) to show the limit. The limit is 3.

Step 1: 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 \( |(5x - 2) - 3| < \varepsilon \).

Step 2: Simplify the Absolute Value Inequality

Simplify \( |5x - 2 - 3| < \varepsilon \).\[ |5x - 2 - 3| = |5x - 5| = 5|x - 1| \].

Step 3: Relate x to \\varepsilon

For \( |5x - 5| < \varepsilon \), we want to express this in terms of \( |x - 1| \):\[ 5|x - 1| < \varepsilon \] implies \[ |x - 1| < \frac{\varepsilon}{5} \].

Step 4: Choose \\delta

Choose \( \delta = \frac{\varepsilon}{5} \). This ensures that whenever \( |x - 1| < \delta \), \(|5x - 5| < \varepsilon\).

Step 5: 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, \( \lim_{x \to 1} (5x - 2) = 3 \).

Key Concepts

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.

When evaluating the limit \( \lim_{x \to 1} (5x - 2) \), we're trying to determine what value \( 5x - 2 \) approaches as \( x \) gets closer to \( 1 \). The use of epsilon-delta definition is a precise way to prove this limit.

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 \( \varepsilon \).

• \( \varepsilon \) 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 \( f(x) \).

In proving \( \lim_{x \to 1} (5x - 2) = 3 \), we use this definition to establish a relationship between \( \varepsilon \) and \( \delta \).

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 \( |x - 1| < \frac{\varepsilon}{5} \).

• 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 \( \delta \).

This inequality provides the necessary condition to choose \( \delta = \frac{\varepsilon}{5} \), ensuring the function remains within \( \varepsilon \) of the limit value.

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:

• Determining \( \varepsilon \) sets the tolerance for output variation.
• Calculating \( \delta \) establishes the input range necessary to keep function values within this tolerance.

Each epsilon-delta proof utilizes similar logical steps, reinforcing the connectivity between input behavior and function trends as highlighted in calculus principles.