Problem 57 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indicated limit, if it exists. \(\lim _{x \rightarrow-2} \frac{x^{2}-x-6}{x^{2}+x-2}\)

Step-by-Step Solution

Answer

The limit of the given rational function as x approaches -2 is: \(\lim_{x \rightarrow -2} \frac{x^2 - x - 6}{x^2 + x - 2} = \frac{-5}{-3}\).

Step 1: Factor the numerator and denominator of the rational function

First, we will factor the numerator and denominator of the given rational function to see if we can simplify it: Numerator: \(x^2 - x - 6\) Find two numbers that multiply to -6 and add to -1. These numbers are -3 and 2. So, the factored form is: \((x -3)(x +2)\) Denominator: \(x^2 + x - 2\) Find two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. So, the factored form is: \((x -1)(x +2)\) Thus, the rational function becomes: \[\frac{(x-3)(x+2)}{(x-1)(x+2)}\]

Step 2: Simplify the rational function

Now, we can simplify the rational function by canceling out common factors in the numerator and denominator. In this case, we see that (x+2) is a common factor: \[\frac{(x-3)(x+2)}{(x-1)(x+2)} = \frac{x-3}{x-1}\]

Step 3: Evaluate the limit

Now that we have simplified the rational function, we can evaluate the limit by plugging in x = -2: \[\lim_{x \rightarrow -2} \frac{x-3}{x-1} = \frac{(-2)-3}{(-2)-1} = \frac{-5}{-3} \]

Step 4: Write the final answer

The limit of the given rational function as x approaches -2 is: \[\lim_{x \rightarrow -2} \frac{x^2 - x - 6}{x^2 + x - 2} = \frac{-5}{-3}\]

Key Concepts

Rational Functions

Rational functions are all about fractions involving polynomials. Think of a rational function as a large fraction where the top part is one polynomial (numerator) and the bottom part is another polynomial (denominator). Learning how they behave, especially as we approach specific points or infinity, is a crucial skill in calculus. It's like understanding how ingredients in a recipe change its taste. Key features of rational functions include:

• Holes: Occur when a factor is canceled from both the numerator and the denominator.
• Asymptotes: These are lines that the graph of the function gets infinitely close to but never actually touches.
• Zeros: Determined by setting the numerator equal to zero, they tell you where the function equals zero.

For evaluating limits, our interest is often in places where the function might not be defined at first glance because the denominator is zero. By factoring polynomials or simplifying expressions, we can often "see through" the indeterminacy and find a clear answer just like in the above exercise.

Factoring Polynomials

Factoring polynomials is like breaking down a big puzzle into smaller, easier-to-solve pieces. It involves writing a polynomial as a product of its simpler factors. This is similar to finding the prime factors of a number in basic arithmetic.Here's how you can approach it:

• Identify the polynomial you want to factor.
• Look for common factors using methods such as trial multiplication, using synthetic division, or applying the quadratic formula.
• In quadratics, look for two numbers that multiply to the constant term (last number) and add to the coefficient of the middle term.

In our example, the numerator \(x^2 - x - 6\) factors into \(x-3\) and \(x+2\). Similarly, the denominator \(x^2 + x - 2\) factors into \(x-1\) and \(x+2\). Factoring helps simplify the expression by revealing common terms or factors, allowing for cancellation which is a key step towards simplification.

Simplifying Expressions

Simplifying expressions involves reducing them to a form that is easier to work with, eliminating any unnecessary complexity. Rather than solving them outright, we make them easier to understand and manage. It's like cleaning your workspace before you start a big project. In a rational function, this process often involves canceling out common factors found in both the numerator and the denominator. In the exercise, after factoring, we noticed that \(x+2\) was a common factor in both parts. By canceling these, we simplified the expression to a much tidier \(\frac{x-3}{x-1}\).Why simplify? Simplification helps us focus only on what's essential and ignore terms that complicate but do not materially affect our work. It makes finding limits and other calculations more straightforward.

Evaluating Limits

Evaluating limits is like predicting the behavior of a function as it gets close to a certain point. This concept is essential in calculus as it forms the basis for understanding continuity, derivatives, and integrals. To evaluate a limit:

• Observe the behavior of the function as it approaches the value from both sides.
• Simplify the expression to handle indeterminate forms such as \(\frac{0}{0}\).
• Substitute the value or use limit laws/rules to determine the approaching value.

In our example, after simplifying the rational function, we plugged in \(x = -2\) into the simplified expression \(\frac{x-3}{x-1}\). This gave us a result of \(\frac{-5}{-3}\), since the simplification removed the indeterminate form. Evaluating limits often requires these steps, making it easier to solve problems accurately and efficiently.