Problem 35 - Precalculus: Mathematical for Calculus

Question

(a) What is wrong with the following equation? $$ \frac{x^{2}+x-6}{x-2}=x+3 $$ (b) In view of part (a), explain why the equation $$ \lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3) $$ is correct.

Step-by-Step Solution

Answer

The first equation is incorrect due to a domain restriction at \(x = 2\). The second equation is correct as limits consider behavior approaching \(x = 2\) but not at it.

Step 1: Simplify the Expression

First, we need to factor the numerator of the given equation. The expression \(x^2 + x - 6\) can be factored as \((x - 2)(x + 3)\). So, the equation becomes: \[\frac{(x - 2)(x + 3)}{x-2} = x + 3\] The incorrect assumption is that \((x - 2)\) will cancel out completely without consideration. This assumption does not take into account the domain of the original expression.

Step 2: Identify the Domain

The original expression \(\frac{x^2+x-6}{x-2}\) has a restriction: \(x eq 2\), because substituting \(x = 2\) makes the denominator zero, which is undefined. This restriction means that while simplification allows us to write \(x + 3\), the simplified form is not equivalent for \(x = 2\). Therefore, the equation \(\frac{x^2+x-6}{x-2} = x+3\) is incorrect since it implies equality for all \(x\), including \(x = 2\).

Step 3: Verify the Limit Equation

The limit expression given is:\[\lim _{x \rightarrow 2} \frac{x^2 + x - 6}{x - 2} = \lim _{x \rightarrow 2}(x + 3)\]To solve this, realize that as \(x\) approaches 2, the expression \((x - 2)\) exists in both numerator and denominator so we can 'cancel' it under the limit context. Effectively, we can consider:\[\lim _{x \rightarrow 2}(x + 3) = 5\].This equation is correct because limits evaluate the behavior as \(x\) approaches the value rather than at the value itself.

Key Concepts

Rational Functions

A rational function is any function that can be written as a quotient of two polynomial expressions. Consider the function \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). The important aspect of rational functions is that their domain excludes any values that make the denominator zero.
This can lead to important behaviors at these points, including vertical asymptotes or holes in the graph.When examining the equation \(\frac{x^2 + x - 6}{x-2} = x+3\), we encounter a rational expression. The misunderstanding occurs when simplifying, not considering that \((x-2)\) could be zero, leading to terms that cancel out improperly. Simplifying correctly requires considering the domain of the function beforehand.
Simplification without domain consideration often leads to incorrect assumptions about equivalence around undefined points.

Domain of a Function

Understanding the domain of a function is crucial for solving and simplifying functions correctly. The domain of a function comprises all the possible input values \(x\) for the function to produce a valid output. This is particularly important when dealing with rational functions, as division by zero is undefined and must be avoided.For the function \(\frac{x^2 + x - 6}{x-2}\), the value \(x = 2\) is excluded from the domain as it causes the denominator to be zero. Defining the domain as all real numbers except \(x = 2\) helps prevent logical errors during simplification or analysis.
In solving limits or evaluating rational equations, explicitly recognizing these excluded points ensures correct mathematical interpretation and computation.

Factoring Polynomials

Factoring polynomials is a fundamental skill in algebra and precalculus. It involves rewriting a polynomial as a product of simpler expressions, often to simplify complex algebraic fractions.
For the polynomial expression \(x^2 + x - 6\), it can be factored into \((x - 2)(x + 3)\). This factorization tells us that \(x = 2\) and \(x = -3\) are the roots or solutions when the polynomial equals zero.Using this factorization, you can simplify rational expressions by canceling common factors in the numerator and denominator, provided the common factor does not make the denominator zero.
Understanding how to factor polynomials effectively allows for easier computation of limits and solving of rational functions, while respecting the domain restrictions inherently present.