Problem 31 - 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 \(x\) $$ 0<|x-c|<\delta \quad \Rightarrow \quad|f(x)-L|<\epsilon $$ $$ f(x)=3-2 x, \quad c=3, \quad \epsilon=0.02 $$

Step-by-Step Solution

Answer

Limit is -3; choose \( \delta = 0.01 \).

Step 1: Calculate the Limit

Identify the function, point, and calculate the limit. The function given is \( f(x) = 3 - 2x \), and we need to find \( L = \lim_{x \to 3} f(x) \). By substituting \( x = 3 \) into \( f(x) \), we have \( f(3) = 3 - 2(3) = 3 - 6 = -3 \). Thus, \( L = -3 \).

Step 2: Set Up the Inequality for δ

We need to find a \( \delta > 0 \) such that for all \( x \), if \( 0 < |x - 3| < \delta \), then \( |f(x) + 3| < 0.02 \). Starting with \( f(x) = 3 - 2x \), we have \( |3 - 2x + 3| < 0.02 \), which simplifies to \( |-2x + 6| < 0.02 \).

Step 3: Simplify the Inequality

The inequality \( |-2x + 6| < 0.02 \) can be rewritten as \( 2|x - 3| < 0.02 \). Dividing both sides by 2 gives \( |x - 3| < 0.01 \). Thus, \( |x - 3| < \delta = 0.01 \).

Step 4: Conclusion

We have calculated the limit as \( L = -3 \) and determined that \( \delta = 0.01 \) satisfies the condition that for all \( x \) where \( 0 < |x - 3| < \delta \), \( |f(x) + 3| < 0.02 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition of a limit is a formal and precise way of stating what it means for a function to approach a certain value as its inputs approach a certain point. Here’s how it works in simple terms:

• Epsilon (\( \epsilon \)): This is the small positive number that represents the distance our function's value can be from the limit, denoted as \( L \). This means we want our function's outputs to be inside a 'window' of \( L \pm \epsilon \).
• Delta (\( \delta \)): This is another small positive number, representing how close \( x \) can get to the point \( c \) without actually being equal to \( c \), such that the function's values remain within the epsilon window.

The core idea is: no matter how small we make \( \epsilon \), we can find a \( \delta \) such that any \( x \) within \( \delta \) of \( c \), except \( c \) itself, forces the function's value to be within \( \epsilon \) of \( L \). This ensures the function truly gets close to \( L \) as \( x \) gets close to \( c \). In the problem above, we find \( \delta \) to ensure the function's limit behavior is as expected within the given epsilon window.

Continuous Functions

Continuous functions are those that have no breaks, jumps, or holes in their graphs. Think of them as functions where you can draw their entire graph without lifting your pencil.

• A function \( f(x) \) is continuous at a point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) equals \( f(c) \). This means there is no sudden change in the value of \( f(x) \) when \( x \) is exactly \( c \).
• If a function is continuous everywhere in its domain, it is said to be a continuous function.

The function provided, \( f(x) = 3 - 2x \), is linear and thus continuous everywhere. This means at any point \( c \), including \( c = 3 \), the limit of \( f(x) \) as \( x \) approaches \( c \) is exactly \( f(c) \). In our exercise, this reliability of link between \( c \) and \( f(c) \) simplifies finding limits because continuity assures us the function's behavior will be steady and predictable.

Limit of a Function

The limit of a function at a certain point describes the behavior of the function as the input values get arbitrarily close to a given point. It essentially provides the destination of the function's values as the input approaches a specific number, without necessarily being equal to that number.To find the limit, we follow these steps:

• Identify the function \( f(x) \), the point \( c \) we are approaching, and any constants such as \( \epsilon \).
• Substitute the point \( c \) into the function \( f(x) \) to simplify calculations and hint at what the limit \( L \) might be.
• Verify the limit by considering the epsilon-delta definition, ensuring the function’s output stays within \( \epsilon \) of \( L \) whenever the input is within \( \delta \) of \( c \), but not equal to \( c \).

In the example, substituting \( x = 3 \) into \( f(x) = 3 - 2x \) revealed that the limit \( L \) is \(-3\). This confirms the function decreases linearly as \( x \) increases, and thus approaches \(-3\) at \( x = 3 \), aligning with our manually calculated limit from earlier.