Problem 22 - Calculus and its applications

Question

The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow 1} \frac{x^{2}+5 x-6}{x^{2}-1} $$

Step-by-Step Solution

Answer

The limit is \(\frac{7}{2}\).

Step 1: Identify the Problem Type

This problem involves finding the limit of a rational function as it approaches a certain value. Notice that directly substituting \(x = 1\) gives the indeterminate form \(\frac{0}{0}\).

Step 2: Factor the Numerator and Denominator

To simplify, factor both the numerator and the denominator. The numerator \(x^2 + 5x - 6\) factors as \((x - 1)(x + 6)\), while the denominator \(x^2 - 1\) factors as \((x - 1)(x + 1)\).

Step 3: Simplify the Expression

Cancel the common term \((x-1)\) in both the numerator and denominator. The simplified expression becomes \(\frac{x + 6}{x + 1}\).

Step 4: Evaluate the Simplified Limit

Now, find the limit of the simplified expression as \(x\) approaches 1. Substitute \(x = 1\) into \(\frac{x + 6}{x + 1}\) to get \(\frac{1 + 6}{1 + 1} = \frac{7}{2}\).

Key Concepts

Rational Functions

Rational functions are expressions that involve fractions where both the numerator and the denominator are polynomials. These functions can often be tricky to handle if you're attempting to directly compute limits, especially when they result in forms like \( \frac{0}{0} \). Rational functions are frequently used in calculus to determine behavior around points of interest, such as limits and continuity. For the given exercise, you have a rational function \( \frac{x^2 + 5x - 6}{x^2 - 1} \). Initially, substituting \( x = 1 \) into this function, you get the indeterminate form \( \frac{0}{0} \). This happens because both the numerator and the denominator evaluate to zero. To overcome this challenge, you need to simplify the function algebraically or use other methods like graphing or creating a table. By breaking down the polynomials, you can gain insight into the behavior of rational functions and handle limits more effectively. Remember, understanding rational functions is essential because they model many real-world phenomena and provide a concrete way to explore how quantities relate to each other.

Indeterminate Forms

Indeterminate forms arise when evaluating limits leads to an uncertain result, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other ambiguous expressions. These forms do not provide enough information to determine the limit directly. Therefore, they require further analysis through algebraic simplification or applying limit theorems.In the exercise, the direct substitution of \( x = 1 \) yields the form \( \frac{0}{0} \), signaling an indeterminate form. This scenario guides you to find alternative methods to evaluate the limit. Rather than concluding that the limit does not exist, you factor the expression or use techniques like L'Hôpital's Rule or graphical analysis. By simplifying the expression to \( \frac{x+6}{x+1} \), you resolve the indeterminate form.Understanding indeterminate forms is crucial in calculus as they frequently appear when dealing with limits of functions at points of discontinuity. With the proper techniques, you can transform an uncertain result into a meaningful value.

Factoring Polynomials

Factoring polynomials is an essential skill in calculus, especially when working with rational functions experiencing limits that result in indeterminate forms. Factoring allows you to simplify expressions, making it possible to cancel common terms in the numerator and the denominator.In this exercise, the polynomial in the numerator \( x^2 + 5x - 6 \) is factorable into \( (x-1)(x+6) \), while the denominator \( x^2 - 1 \) factorizes to \( (x-1)(x+1) \). By expanding the expression through factoring, you identify and cancel the common factor \( (x-1) \). This strategic move reduces the complexity of the rational function, leading to the simplified expression \( \frac{x+6}{x+1} \), which can then be evaluated directly.Mastering polynomial factoring provides the tools you need to handle various algebraic manipulations, leading to correct computations of limits and a broader understanding of functional behavior.