Problem 19 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Either find the given limit or show it does not exist. If the limit is infinite, indicate whether it is \(+\infty\) or \(-\infty\). $$ \lim _{x \rightarrow 1} \frac{x^{2}+x-2}{x^{2}-1} $$

Step-by-Step Solution

Answer

The limit is \(\frac{3}{2}\).

Step 1: Simplify the Expression

First, factorize the numerator and denominator: For the numerator: \(x^2 + x - 2\) can be written as \((x-1)(x+2)\). For the denominator: \(x^2 - 1\) is a difference of squares and can be written as \((x-1)(x+1)\).

Step 2: Rewrite the Limit

After factoring, the limit expression becomes: \[ \frac{(x-1)(x+2)}{(x-1)(x+1)} \]

Step 3: Cancel Common Factors

Cancel the common factor \((x-1)\) in the numerator and the denominator: \[ \frac{(x-1)(x+2)}{(x-1)(x+1)} = \frac{x+2}{x+1} \]

Step 4: Evaluate the Limit

Now, with the simplified expression, substitute \(x\) approaching \(1\): \[ \frac{1+2}{1+1} = \frac{3}{2} \]

Key Concepts

Limit Evaluation

In calculus, evaluating limits helps us understand the behavior of a function as it approaches a specific point. A limit can reveal interesting properties such as continuity, asymptotic behavior, and rates of change.
When dealing with exercises like the one provided, where we need to find the limit of a function, follow these steps:

• First, attempt direct substitution. If this causes indeterminate forms like \(\frac{0}{0}\), use algebraic manipulation.
• Simplify the expression, often through factorization.
• Cancel out common factors if possible.
• Substitute the value back into the simplified expression to find the limit.

In the example given, \(\lim_{{x \rightarrow 1}} \frac{{x^2 + x - 2}}{{x^2 - 1}}\), we follow these steps to ultimately get \(\frac{3}{2}\). This tells us the function \(\frac{{x^2 + x - 2}}{{x^2 - 1}}\) approaches \(\frac{3}{2}\) as \(x\) approaches 1.

Rational Functions

A rational function is a fraction of two polynomials. The key characteristics of rational functions include:

• They can have vertical asymptotes at points where the denominator is zero.
• They may have holes (removable discontinuities) where both numerator and denominator are zero.
• The behavior at infinity is often determined by the degrees of the numerator and denominator.

For the given exercise, the rational function is \(\frac{{x^2 + x - 2}}{{x^2 - 1}}\). Initially, plugging \(x = 1\) into the function results in \(\frac{0}{0}\). This indicates a possible hole rather than an asymptote. By factorizing, we simplify the function to \(\frac{x+2}{x+1}\). This new function helps us find the limit while removing the indeterminate form, revealing that the limit is \(\frac{3}{2}\).

Factorization

Factorization is breaking down a complex expression into simpler factors that, when multiplied together, give the original expression. In calculus, it's crucial for simplifying functions to evaluate limits.
For our example, factorizing both the numerator and the denominator aids in canceling out common terms:

• Numerator: \(x^2 + x - 2 = (x-1)(x+2)\)
• Denominator: \(x^2 - 1 = (x-1)(x+1)\)

By factorizing these expressions, we cancel out the common term \(x-1\), simplifying the limit expression to \(\frac{x+2}{x+1}\). This crucial step moves us from an indeterminate form to an easily evaluated expression. Factorization simplifies complex problems and is a powerful algebraic tool in limit evaluation and beyond.