Problem 33 - 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 \(x\) $$ 0<|x-c|<\delta \quad \Rightarrow \quad|f(x)-L|<\epsilon $$ $$ f(x)=\frac{x^{2}-4}{x-2}, \quad c=2, \quad \epsilon=0.05 $$

Step-by-Step Solution

Answer

The limit is 4 and \( \delta = 0.05 \).

Step 1: Simplify the Function

To find the limit, first simplify \( f(x) = \frac{x^2 - 4}{x-2} \). Notice that the numerator can be factored as \( (x-2)(x+2) \) so the function simplifies to \( f(x) = x+2 \) for \( x eq 2 \).

Step 2: Calculate the Limit

Plug in \( c = 2 \) into the simplified function \( f(x) = x + 2 \). Thus, \( L = \lim_{x \to 2} f(x) = 2 + 2 = 4 \).

Step 3: Understand the \'\(\epsilon\)' Condition

We are given \( \epsilon = 0.05 \). The condition we need to satisfy is \( |f(x) - L| < \epsilon \). Substituting \( L = 4 \), we need \( |x + 2 - 4| < 0.05 \). Simplifying, this becomes \( |x - 2| < 0.05 \).

Step 4: Determine \( \delta \)

From the inequality \( |x - 2| < 0.05 \), we observe that the smallest such \( \delta \) satisfying the condition is 0.05. Thus, \( \delta = 0.05 \). The condition \( 0 < |x-2| < \delta \) implies \( |f(x) - 4| < 0.05 \) for \( \delta = 0.05 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a rigorous method used to define the limit of a function. It's about understanding how close we can get to a function's limit by controlling the distance between the values of the function and the limit itself. The approach uses two Greek letters: \( \epsilon \) and \( \delta \). Here’s how it works:

• \( \epsilon \) (epsilon) represents how close you want the function's value to be to the limit.
• \( \delta \) (delta) is the distance we're allowed to move away from a certain point \( c \) on the \( x \)-axis.

The goal is to make \(|f(x) - L| < \epsilon\) whenever \(0 < |x - c| < \delta\). This means, if \( x \) is within \( \delta \) units of \( c \), then \( f(x) \) will be within \( \epsilon \) units of \( L \), the limit as \( x \) approaches \( c \). This definition ensures that we can make \( f(x) \) as close to \( L \) as desired by choosing sufficiently small values for \( \delta \). In practice, it’s like setting a tolerance for how precise we want the approximation to be.When you solve a limit problem using the epsilon-delta definition, you essentially showcase that for a specific \( \epsilon \), you can always find a corresponding \( \delta \) that satisfies the condition. It's a fundamental concept in calculus that develops the precision needed for defining limits formally.

Simplifying Rational Functions

Simplifying rational functions is a crucial step in finding limits or solving equations where functions can be expressed as fractions of polynomials. A rational function is of the form \( \frac{P(x)}{Q(x)} \) where both \( P(x) \) and \( Q(x) \) are polynomials. Here's how you go about simplifying these functions:

• Identify factors in the numerator and the denominator that are common.
• Cancel out those common factors, but make sure that the domain of the function doesn’t change when removing factors.

In our example, \( f(x) = \frac{x^2 - 4}{x - 2} \), the numerator \( x^2 - 4 \) can be factored into \( (x-2)(x+2) \). We can simplify \( f(x) \) by canceling the common factor \( (x-2) \), yielding \( f(x) = x+2 \), provided \( x eq 2 \). Simplification helps make the problem more manageable, allowing us to solve it efficiently. We must always remember not to lose track of restrictions in the domain; here, \( x eq 2 \) still holds even after simplification.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into the product of simpler polynomials. This technique is beneficial when solving equations or simplifying expressions.

• Begin by examining the polynomial for any common factors among its terms.
• Consider special patterns like the difference of squares, perfect square trinomials, or the sum and difference of cubes.

In the exercise, the polynomial \( x^2 - 4 \) is a classic example of the difference of squares. It can be factored into \( (x-2)(x+2) \). Recognizing such patterns is vital in simplifying expressions and solving equations efficiently. Factoring is not just about finding products; it's about revealing the structure of a polynomial, which aids in manipulation and simplification important steps in calculus and algebra. Remember, practice makes perfect—persistently working on factoring exercises enhances your ability to spot these patterns more quickly and accurately.