Problem 36 - Calculus: Early Transcendentals

Question

Prove that \( \displaystyle \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} \).

Step-by-Step Solution

Answer

The limit \( \lim_{x \to 2} \frac{1}{x} \) is \( \frac{1}{2} \) using the epsilon-delta definition.

Step 1: State the Limit Definition

To prove that \( \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} \), we use the epsilon-delta definition of a limit. This definition states: For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \), then \( \left| \frac{1}{x} - \frac{1}{2} \right| < \varepsilon \).

Step 2: Express the Inequality

We need to manipulate \( \left| \frac{1}{x} - \frac{1}{2} \right| < \varepsilon \) to find \( \delta \). Begin by simplifying: \[ \left| \frac{1}{x} - \frac{1}{2} \right| = \left| \frac{2-x}{2x} \right| = \frac{|2-x|}{2|x|}. \] We want this to be less than \( \varepsilon \).

Step 3: Find the Bound for x

Assume \( |x - 2| < 1 \), which implies \( 1 < x < 3 \). In this interval, \( |x| > 1 \), so \( 2|x| > 2 \). This gives \( \frac{|2-x|}{2|x|} < \frac{|2-x|}{2}. \)

Step 4: Relate to Epsilon

To ensure \( \frac{|2-x|}{2} < \varepsilon \), choose \( |2-x| < 2\varepsilon \). When \( |x-2| < \min(1, 2\varepsilon) = \delta \), both previous conditions hold.

Step 5: Conclude the Proof

By choosing \( \delta = \min(1, 2\varepsilon) \), we have shown that whenever \( 0 < |x - 2| < \delta \), it follows that \( \left| \frac{1}{x} - \frac{1}{2} \right| < \varepsilon \). Thus, \( \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} \) is proven.

Key Concepts

Inequality Manipulation

Manipulating inequalities is a crucial step when proving limits using the epsilon-delta definition. In this proof, we deal with the inequality

• \( \left| \frac{1}{x} - \frac{1}{2} \right| < \varepsilon \).

First, we want to express this inequality in a form that allows us to find a suitable \( \delta \). This involves simplifying it to find a relationship between \( |x - 2| \) and \( \varepsilon \).

We rewrite the difference between \( \frac{1}{x} \) and \( \frac{1}{2} \) as

• \( \left| \frac{2-x}{2x} \right| = \frac{|2-x|}{2|x|} \).

This expression highlights how \( |x - 2| \) controls the difference we are interested in. We then proceed by estimating \( |x| \) and managing the fractions to ensure our inequalities remain valid.

The trick lies in estimating \( |x| \) within a restricted interval, leading to the relaxed inequality \( \frac{|2-x|}{2|x|} < \frac{|2-x|}{2} \). This simplification allows us to control the expression further by relating it directly to \( \varepsilon \). Exploiting this manipulation is a common tactic when handling absolute values in limit proofs.

Limit Proof

A limit proof involves demonstrating that a function approaches a specific value as the input approaches a particular point. To prove a limit rigorously, we rely on definitions, such as the epsilon-delta definition. In this problem, the goal is to show:

• \( \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} \).

This requires us to ensure that for every small chosen \( \varepsilon > 0 \), there exists a corresponding \( \delta > 0 \) such that

• \( 0 < |x - 2| < \delta \) implies \( \left| \frac{1}{x} - \frac{1}{2} \right| < \varepsilon \).

We accomplish this by manipulating inequalities to express \( |x - 2| \) in terms of \( \varepsilon \).

By bounding \( |x| \) through assumptions like \( |x - 2| < 1 \), we simplify the relationships further to connect them to our \( \varepsilon \).
This systematic process culminates in choosing a successful \( \delta \). The final \( \delta \), often found as the minimum of multiple bounds, guarantees our target inequality holds for any tiny \( \varepsilon \).

This kind of proof solidifies our understanding and application of limit concepts.

Epsilon-Delta Technique

The epsilon-delta technique is fundamental in understanding limits in calculus. It formalizes the intuitive notion of "closeness" by defining the limit \( \lim_{x \to a} f(x) = L \) with precise language:

• For every \( \varepsilon > 0 \), there is a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \varepsilon \).

In practical terms, this involves constructing a scenario where manipulating \( |x - a| \) smaller than \( \delta \) forces \( |f(x) - L| \) to be smaller than \( \varepsilon \).

The main challenge is choosing \( \delta \) correctly, often by manipulating inequalities. In our specific problem, we used the expression

• \( \frac{|2-x|}{2} < \varepsilon \)

to define \( \delta = \min(1, 2\varepsilon) \). By design, \( \delta \) accounts for uncertainties and ensures the condition holds for any \( x \) approaching the limit point.

The epsilon-delta process, although appearing complex initially, becomes robust with practice. It reinforces the precision required in calculus, ensuring every limit defined is mathematically sound.