Problem 10 - Single Variable Calculus

Question

You are given \(\lim _{x \rightarrow a} f(x)=L\) and a tolerance \(\varepsilon\). Find a number \(\delta\) such that \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\) \(\lim _{x \rightarrow 2} \frac{1}{x}=\frac{1}{2} ; \quad \varepsilon=0.05\)

Step-by-Step Solution

Answer

We find that \(\delta = \frac{2}{11}\) is a suitable choice, as it ensures that \(|\frac{1}{x} - \frac{1}{2}| < 0.05\) when \(0 < |x - 2| < \frac{2}{11}\).

Step 1: 1. Write down the main components of the problem

We are given the limit of a function, the value of the function, and a tolerance. Our goal is to find a number \(\delta\). Here, we have: Function: \(f(x) = \frac{1}{x}\) \(a = 2\) \(L = \frac{1}{2}\) \(\varepsilon = 0.05\)

Step 2: 2. Rewrite the definition of a limit using given information

Using the given information, rewrite the definition of a limit: \(|f(x) - L| < \varepsilon\) whenever \(0 < |x - a| < \delta\) Substitute the given values: \(|\frac{1}{x} - \frac{1}{2}| < 0.05\) whenever \(0 < |x - 2| < \delta\)

Step 3: 3. Manipulate the inequality to isolate \(\delta\)

We want to find a condition for \(\delta\) that depends on \(|x - 2|\): \(|\frac{1}{x} - \frac{1}{2}| < 0.05\) Find a common denominator and simplify: \(|\frac{2 - x}{2x}| < 0.05\) As \(x\) approaches \(2\), the term \(|2 - x|\) in the numerator is always positive. So we can rewrite the inequality as: \(\frac{2 - x}{2x} < 0.05\) Now, isolate \(\delta = |x - 2|\): \(\frac{\delta}{2(2 - \delta)} < 0.05\)

Step 4: 4. Solve for \(\delta\)

Now we need to find a value for \(\delta\) that satisfies the inequality: \(\frac{\delta}{2(2 - \delta)} < 0.05\) First, we can multiply both sides by \(2(2 - \delta)\), obtaining: \(\delta < 0.1(2 - \delta)\) Now solve for \(\delta\): \(\delta < 0.2 - 0.1\delta\) \(\delta + 0.1\delta < 0.2\) \(\delta(1 + 0.1) < 0.2\) \(\delta < \frac{0.2}{1.1}\) \(\delta < \frac{2}{11}\) So, we can choose \(\delta = \frac{2}{11}\). This means that if \(0 < |x - 2| < \frac{2}{11}\), then \(|\frac{1}{x} - \frac{1}{2}| < 0.05\).

Key Concepts

Limit of a Function

In calculus, the concept of a limit is fundamental. It helps us understand how a function behaves as the input value approaches a particular point. When we say the limit of a function \( f(x) \) as \( x \) approaches \( a \) is \( L \), denoted \( \lim_{x \rightarrow a} f(x) = L \), it means the function values get arbitrarily close to \( L \) as \( x \) gets closer to \( a \).

This intuitive idea is crucial for understanding concepts like continuity and differentiability. In the given exercise, we have the limit \( \lim_{x \rightarrow 2} \frac{1}{x} = \frac{1}{2} \). It tells us that as \( x \) gets near to \( 2 \), the value of \( \frac{1}{x} \) should get closer to \( \frac{1}{2} \).

Understanding limits is essential for dealing with all sorts of calculus problems, ranging from evaluating derivatives to integrating complex functions.

Epsilon-Delta Proof

The epsilon-delta (\( \varepsilon-\delta \)) definition of a limit provides a rigorous way to define what it means for a limit to exist. For a function \( f(x) \), we say the limit as \( x \) approaches \( a \) is \( L \) (i.e., \( \lim_{x \rightarrow a} f(x) = L \)) if for every small positive number \( \varepsilon \) (epsilon), there is a corresponding small positive number \( \delta \) (delta) such that whenever \( 0 < |x - a| < \delta \), it guarantees \(|f(x) - L| < \varepsilon\).

In other words, no matter how close you want \( f(x) \) to be to \( L \), you can always find an interval around \( a \) (but excluding \( a \) itself) that keeps \( f(x) \) that close to \( L \). Developing an intuition for epsilons and deltas is fundamental to mastering limits. It offers a way to handle complex mathematical problems with precision.

In our problem, we were asked to find such a \( \delta \) that works for \( \varepsilon = 0.05 \), ensuring \(|\frac{1}{x} - \frac{1}{2}| < 0.05\) when \( 0 < |x - 2| < \delta \). Through manipulation, we determined \( \delta = \frac{2}{11} \).

Calculus Problem Solving

Solving calculus problems often requires both algebraic manipulation and conceptual understanding. Given a limit problem, you need to apply the concepts of limits and use algebra to isolate and determine values that satisfy the given conditions.

In the provided exercise, we needed to show that for the limit \( \lim_{x \rightarrow 2} \frac{1}{x} = \frac{1}{2} \), there exists a \( \delta \) such that \( |\frac{1}{x} - \frac{1}{2}| < 0.05 \) whenever \( 0 < |x - 2| < \delta \). We approached this by:

• Understanding the given function and limit values.
• Rewriting the problem using epsilon-delta criteria.
• Engaging in algebraic manipulation to isolate \( \delta \).
• Solving the inequality to find the suitable \( \delta \).

This kind of problem-solving skill is crucial for tackling not only limits but other calculus topics like continuity, derivatives, and integrals. Practice helps you become proficient in manipulating equations and understanding the underlying concepts.