Problem 10 - Calculus Early Transcendentals

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 equation is undefined at \( x = 2 \) due to division by zero. The limit statement circumvents this by considering approaching values, making the result valid: \( \lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2} = 5 \).

Step 1: Identify the Error in the Equation

The equation \( \frac{x^{2}+x-6}{x-2}=x+3 \) is incorrect as a standard algebraic equation because the left-hand side is undefined when \( x = 2 \). Substituting \( x = 2 \) into the denominator makes the expression \( \frac{0}{0} \), which is undefined.

Step 2: Examine the Factorization

To understand the behavior of the expression \( \frac{x^{2}+x-6}{x-2} \), factor the numerator: \[ x^2 + x - 6 = (x-2)(x+3) \] Therefore, the original expression simplifies to \( \frac{(x-2)(x+3)}{x-2} \). This simplifies to \( x+3 \) for \( x eq 2 \).

Step 3: Explain Why the Limit Statement is Correct

The given limit \( \lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3) \) is correct. The reason is that as \( x \) approaches 2, we can cancel the \( x-2 \) terms, leaving \( x+3 \), which is continuous and well-defined for \( x = 2 \). Thus, the limit of the expression as \( x \) approaches 2 is the same as evaluating \( x+3 \) at \( x = 2 \), which is 5.

Step 4: Evaluate the Limit

To find \( \lim _{x \rightarrow 2}(x+3) \), directly substitute \( x = 2 \). This gives: \[ x + 3 = 2 + 3 = 5 \] Thus, the limit is 5.

Key Concepts

Rational Expressions

When working with rational expressions, we deal with fractions where both the numerator and the denominator are polynomials. These types of expressions are widespread in calculus and algebra. A rational expression resembles a fraction, and they require certain conditions to be defined properly.

For instance, to determine where a rational expression is undefined, evaluate the denominator. If it equals zero at any point, the expression will be undefined there. This often happens when the denominator contains variables, which are typical in rational expressions.

• Examine both the numerator and the denominator to comprehensively understand the expression's behavior.
• Identifying the roots of the denominator's polynomial helps in understanding undefined points or points of discontinuity.

Understanding these expressions fully is crucial as it influences further algebraic manipulation and calculus operations such as solving limits.

Undefined Expressions

Undefined expressions are essential to understand because they appear frequently in math, especially when dealing with fractions. An expression becomes undefined when the denominator equals zero, as in the expression \( \frac{x^2+x-6}{x-2} \). Placing \( x=2 \) into the denominator yields \( 0 \), creating an undefined state of \( \frac{0}{0} \).

When terms in a rational expression are zeroed, it signals a necessity for simplification or a different approach, such as finding limits. Identifying and analyzing undefined expressions helps to discern where algebraic simplification could resolve issues before further steps in calculus are taken.

• Always check the denominator for zeros to find points of undefined expressions.
• Recognize that dividing zero by zero isn’t valid algebraically; more analysis, like factoring and limit evaluation, is needed.

Thus, understanding where expressions are undefined is crucial in preventing errors in mathematical reasoning.

Limit Evaluation

Limit evaluation is a process in calculus used to determine the value that a function approaches as the input approaches some value. In the example given, evaluating \( \lim _{x \rightarrow 2} \frac{x^2+x-6}{x-2} \) seems problematic due to the undefined expression at \( x=2 \). However, through algebraic simplification, we can resolve this.

When dealing with limits, focus instead on the behavior of the function around the specified point, rather than directly at it, if it's undefined. In this case, after simplification by canceling the common terms, the limit becomes \( \lim _{x \rightarrow 2}(x+3) \), which can then be evaluated easily.

• Limits help in determining the behavior of functions near points where they could become undefined.
• They are fundamental in dealing with discontinuities in functions.

Evaluating such limits ensures a more accurate understanding of function behavior where direct substitution fails.

Algebraic Simplification

Algebraic simplification is a key skill in mathematics allowing expressions to be transformed to a more workable form. This often involves factoring and reducing expressions, as seen with \( \frac{x^2+x-6}{x-2} \). The first step involves factoring the polynomial: \( x^2 + x - 6 = (x-2)(x+3) \).

Once factored, cancel out the common factors in the numerator and denominator that cause the zero in the denominator, simplifying \( \frac{(x-2)(x+3)}{x-2} \) to just \( x+3 \) for \( x eq 2 \). This simplification is pivotal as it eliminates the undefined nature of the expression for purposes such as limit evaluation.

• Polynomials are often factored to uncover critical points and simplify expressions.
• Cancel common factors to simplify troublesome expressions.

Such algebraic techniques are instrumental in making calculus computations more manageable, especially when dealing with problematic or undefined terms.