Problem 39 - Calculus Early Transcendentals

Question

Evaluate the following limits, where a and \(b\) are fixed real numbers. \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}\)

Step-by-Step Solution

Answer

Question: Evaluate the limit of the given expression: \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}\) Answer: The limit of the given expression as x approaches 1 is 2.

Step 1: Factor the numerator

Factor the numerator to simplify the expression. In this case, the numerator is a difference of squares, so it factors as \((x+1)(x-1)\). Thus, the expression can be rewritten as: \(\frac{(x+1)(x-1)}{x-1}\)

Step 2: Simplify the expression

Now that the expression is factored, we can notice that the factors \((x-1)\) in the numerator and the denominator can cancel each other out. This leaves us with a simplified expression: \(x+1\)

Step 3: Evaluate the limit

With the simplified expression, we can now evaluate the limit as x approaches 1: \(\lim _{x \rightarrow 1} (x + 1)\) As x approaches 1, the value of the expression becomes: \(1 + 1 = 2\) Therefore, the limit is: \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} = 2\)

Key Concepts

Limits in Calculus

The concept of a limit is foundational in calculus, serving as a tool to describe the behavior of functions as they approach a particular point or infinity. When we say we're evaluating the limit of a function as x approaches a specific value, we are not concerned with the function's value at that point, but rather what value it gets closer to as x gets arbitrarily close to that number.

For example, when we evaluate \( \textstyle\frac{x^{2}-1}{x-1} \) at the limit as x approaches 1, it seems problematic since the denominator becomes zero and the expression is undefined. However, limits allow us to understand the value that the function is approaching despite that apparent indeterminacy.

One common technique to evaluate limits involves simplifying the function so that we can substitute the value directly without creating a situation where we divide by zero. This is often done using algebraic manipulations such as factoring polynomials or canceling common factors, which leads us into the other engaging concepts like factoring polynomials and the difference of squares.

Factoring Polynomials

Factoring polynomials, a vital skill in algebra, unlocks the ability to simplify complex expressions and solve equations. At its core, factoring converts a polynomial into a product of its simplest pieces or 'factors', much like finding the prime factorization of a number.

When facing a polynomial, identify if it has common factors in its terms, use formulas such as the difference of squares, or apply the quadratic formula if needed. In the limit we're examining, the quadratic polynomial \( x^{2}-1 \) is a prime candidate for factoring because it is a difference of squares.

It's important when factoring to remember that every polynomial can be expressed uniquely as a product of irreducible polynomials that cannot themselves be factored further over the integers. Becoming comfortable with factoring polynomials is essential when simplifying expressions for evaluating limits in calculus, as seen in this exercise.

Difference of Squares

The 'difference of squares' is a specific form of factoring that applies to expressions where one term is squared and then subtracted from another squared term. The formula can be written as \( a^2 - b^2 = (a+b)(a-b) \). This technique is frequently used in calculus to transform and simplify limits, allowing for the potential cancellation of similar terms.

Understanding and recognizing this pattern is crucial since it produces two linear factors from a quadratic expression, which makes further manipulation and simplification possible. As shown in the provided exercise, the difference of squares methodology allows us to eliminate the indeterminate form by canceling out the same (x-1) factor from the numerator and denominator. After simplifying, we can directly substitute the approaching value into the remaining expression to find the limit, which, in this case, is 2. This illustrates how interconnected and vital these algebraic tools are in the study of calculus.