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Chapter 13: Problem 35

Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 2} \frac{x^{2}-x-2}{x-2}\)

Short Answer

Expert verified
The limit exists and is 3.

Step by step solution

01

Factor the Numerator

The expression given is \( \frac{x^2 - x - 2}{x-2} \). The numerator can be factored. The quadratic \( x^2 - x - 2 \) can be rewritten as \( (x-2)(x+1) \).
02

Simplify the Expression

After factoring the numerator, the expression becomes \( \frac{(x-2)(x+1)}{x-2} \). This simplifies to \( x+1 \) for all \( x eq 2 \).
03

Create a Value Table

Create a table with values of \( x \) approaching 2 from both sides (e.g., 1.9, 1.99, 2.01, 2.1). Calculate the corresponding \( x+1 \) values (\( 2.9, 2.99, 3.01, 3.1 \)).
04

Evaluate the Limit

As \( x \) approaches 2, both from the left and the right, the values of \( x+1 \) approach 3. This suggests that \( \lim_{x \to 2} (x+1) = 3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
In calculus, factoring quadratics is a valuable skill when working with limits. It's often the first step in solving rational functions with limit problems. A quadratic expression has the form \( ax^2 + bx + c \). When factoring, you want to express it as a product of two binomials.
  • Look for two numbers that multiply to the constant term \( c \) and add to the linear coefficient \( b \).
  • For example, with \( x^2 - x - 2 \), we seek two numbers that multiply to \(-2\) and add up to \(-1\).
  • The numbers \(-2\) and \(1\) do exactly this, so we can factor the expression into \((x-2)(x+1)\).
Factoring helps reveal potential simplifications, especially when a potential zero in the denominator is involved. Knowing how to factor simplifies calculations and eases the evaluation of limits.
Rational Functions
Rational functions are ratios of polynomials \( \frac{p(x)}{q(x)} \). They are essential in learning limits as they often present indeterminate forms like \( \frac{0}{0} \). This means direct substitution might not work right away for finding limits.
  • Simplifying the expression is crucial. This often involves canceling out common factors between the numerator and the denominator.
  • In our example, after factoring, we simplify \( \frac{(x-2)(x+1)}{x-2} \) to \( x+1 \), by removing the common term \((x-2)\).
By focusing on the simplified function, evaluating the limit becomes straightforward and saves you from undefined expressions. Rational functions teach an important lesson on the power and necessity of algebraic manipulation in calculus.
Limit Evaluation Steps
Understanding how to evaluate limits involves several steps and concepts. The main goal is to determine the behavior of a function as the input approaches a specific value. Follow these standard steps for limit evaluation:
  • Direct Substitution: First, substitute the target value into the function. If no indeterminate form arises, this is the limit.
  • Factorization & Simplification: If you encounter \( \frac{0}{0} \), factor and simplify the expression like we did in the example \( \frac{x^2-x-2}{x-2} \) becoming \( x+1 \).
  • Numerical Approach: Use values close to the target, such as with a table, to confirm the trend. Pick values like 1.9, 1.99, and check the output values approach a single limit.
  • Graphical Analysis: If possible, visualize the function. A graph helps see how the function behaves as it nears the point of interest.
The combination of these steps allows for a comprehensive understanding of limits, preparing you for more complex calculus problems. Whether through algebraic simplification or numerical tables, these methods make limit evaluation a logical and systematic process.

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