Problem 36 - Thomas Calculus

Question

Each of Exercises \(31-36\) gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim _{x \rightarrow c} f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$ 0<|x-c|<\delta \quad \Rightarrow \quad|f(x)-L|<\epsilon $$ $$ f(x)=4 / x, \quad c=2, \quad \epsilon=0.4 $$

Step-by-Step Solution

Answer

\(L = 2\), \(\delta = \frac{1}{3}\)

Step 1: Identify the Limit

First, substitute the provided value of \( c = 2 \) into the function \( f(x) = \frac{4}{x} \) to compute the limit: \( L = \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{4}{x} = \frac{4}{2} = 2 \). So, \( L = 2 \).

Step 2: Set Up the Inequality

We need \(|f(x) - L| < \epsilon\). Substitute \( f(x) \), \( L \), and \( \epsilon \) into the inequality to get: \[ \left| \frac{4}{x} - 2 \right| < 0.4 \]

Step 3: Solve the Inequality for \(x\)

Simplify the inequality: \(|\frac{4}{x} - 2| < 0.4\) becomes \[-0.4 < \frac{4}{x} - 2 < 0.4\] This results in two inequalities: \( 1.6 < \frac{4}{x} < 2.4 \). Solve each inequality: \(x > \frac{4}{2.4} \) and \( x < \frac{4}{1.6} \). Calculating, we get \( x > \frac{5}{3} \) and \( x < 2.5 \).

Step 4: Determine \( \delta \)

Choose \( \delta \) as the smaller distance of the critical limits from \( c = 2 \). The distances are \(\frac{2 - \frac{5}{3}}{3} \) and \(2.5 - 2\), which gives \(\frac{1}{3}\) and \(0.5\). Hence, \( \delta \) can be chosen as \(\frac{1}{3}\).

Key Concepts

Epsilon-Delta Definition of a Limit

In calculus, the epsilon-delta definition is a formal way to define the limit of a function. It states that for any given positive number \( \epsilon \), there exists a positive number \( \delta \) such that if the distance from \( x \) to \( c \) is less than \( \delta \), then the distance from \( f(x) \) to \( L \) is less than \( \epsilon \). This can be represented as:\[ 0 < |x - c| < \delta \quad \Rightarrow \quad |f(x) - L| < \epsilon \]Let's break this down:

• \( \epsilon \) represents how close we want \( f(x) \) to be to \( L \).
• \( \delta \) is the distance that \( x \) can be from \( c \).
• The expression \( |x - c| \) is the distance between \( x \) and the value \( c \) where the limit is being evaluated.
• When the above condition is satisfied, it tells us that \( f(x) \) is effectively approaching \( L \) as \( x \) approaches \( c \).

The epsilon-delta definition is a precise formulation that underpins the entire concept of limits, ensuring that they work consistently across all types of functions and situations.

Solving Inequalities

Solving inequalities forms a crucial part of analyzing limits, especially when verifying conditions of the epsilon-delta definition. When you set up an inequality like \( |f(x) - L| < \epsilon \), the task is to manipulate this expression to find suitable boundaries for \( x \).Let's look at how we solve inequalities using the example from the exercise:

• You start with an inequality \( \left| \frac{4}{x} - 2 \right| < 0.4 \).
• You break it into two separate inequalities: \( -0.4 < \frac{4}{x} - 2 < 0.4 \).
• Now, solve them separately: \( \frac{4}{x} > 1.6 \) and \( \frac{4}{x} < 2.4 \).
• This yields \( x > \frac{5}{3} \) and \( x < 2.5 \).

Remember that when solving inequalities:

• Perform similar steps as with equations, but be cautious about reversing inequalities when multiplying or dividing by negative numbers.
• Finding where the function is sandwiched between two bounds gives a range that satisfies the condition.
• This range helps to determine \( \delta \).

Finding Delta in Limits

After establishing inequalities, the next step is determining the right \( \delta \) for the limit condition. \( \delta \) represents how close \( x \) needs to be to \( c \) to ensure \( f(x) \) stays close to \( L \).With our exercise:

• We've determined that \( x > \frac{5}{3} \) and \( x < 2.5 \), setting boundaries around \( c = 2 \).
• To find \( \delta \), consider the closer of these boundaries to \( c = 2 \).
• The potential \( \delta \) values are calculated from distances: \( 2 - \frac{5}{3} \) and \( 0.5 \) from \( x < 2.5 \).
• Choosing the smaller distance ensures that for \( 0 < |x - 2| < \delta \), the condition on \( |f(x) - L| < \epsilon \) will hold.
• Here, \( \delta = \frac{1}{3} \) is selected, being the smaller of the two possible distances.

Finding \( \delta \) ensures the limit definition works correctly, indicating that no matter how \( \epsilon \) is chosen, a corresponding \( \delta \) can be found to fit the definition.