Chapter 2: Problem 189
In the following exercises, use the precise definition of limit to prove the given limits. $$\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=6$$
Short Answer
Expert verified
The limit is proven using \( \delta = \epsilon \).
Step by step solution
01
Understanding the limit statement
The limit statement given is \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6 \). This means as \( x \) approaches 3, the value of the function \( \frac{x^2 - 9}{x - 3} \) approaches 6. We will use the precise (epsilon-delta) definition of limit to prove this.
02
Simplifying the function
First, simplify the expression \( \frac{x^2 - 9}{x - 3} \). Notice that \( x^2 - 9 \) can be factored as \( (x - 3)(x + 3) \). Hence, the expression simplifies to \( x + 3 \) for \( x eq 3 \).
03
Setting up the epsilon-delta definition
The definition states that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - 3| < \delta \), it follows that \( |f(x) - L| < \epsilon \), where \( L = 6 \) for this problem.
04
Applying the epsilon-delta definition
Since the simplified function is \( x + 3 \), compute \( |(x + 3) - 6| = |x - 3| \). We want \( |x - 3| < \epsilon \). Thus, we can choose \( \delta = \epsilon \).
05
Concluding the proof
Recognizing that by choosing \( \delta = \epsilon \), the condition \( |x - 3| < \delta \) ensures \( |f(x) - 6| < \epsilon \). Hence, by the epsilon-delta definition, \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6 \) is proven.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Epsilon-Delta Criterion
The epsilon-delta criterion is a formal method to prove limits in calculus. It's a precise way to show that a function approaches a certain value as the input gets closer to a given point. Think of epsilon (\( \epsilon \)) as how close we want the function's value (\( f(x) \)) to be to the limit (\( L \)). Meanwhile, delta (\( \delta \)) is how close \( x \)) needs to be to the target point for this to happen. In practical terms:
- For every small distance you choose for the function's value (\( \epsilon \)), we determine a distance for \( x \)) (\( \delta \)).
- Whenever \( 0 < |x - 3| < \delta \), the condition \( |f(x) - 6| < \epsilon \) is satisfied.
Limit of a Function
The concept of a limit is central to understanding calculus. A limit asks what value a function approaches as its input nears some point. In our exercise, the limit problem asks us to find what value \( \frac{x^2 - 9}{x - 3} \) approaches as \( x \rightarrow 3 \).To determine the limit:
- We analyze how the function behaves around the point, usually including simplification if possible.
- The solution involves assuming the function is uninterruptedly closing in on a specific number, even if it cannot be evaluated directly at the point due to undefined operations like division by zero.
Factoring Polynomials
Factoring polynomials is a basic yet powerful algebraic technique. It converts expressions into a product of simpler forms or factors. This makes complex expressions easier to handle. In our exercise, factoring helps us simplify \( x^2 - 9 \).Here's how it works:
- Identify patterns or use formulas like \( a^2 - b^2 = (a-b)(a+b) \) — a very useful method in such cases.
- Apply the formula to transform \( x^2 - 9 \) into \( (x-3)(x+3) \).
- By canceling common factors with the denominator \( x - 3 \), you simplify the initial expression, making calculations manageable.