Problem 21 - Precalculus: Mathematics for Calculus

Question

Finding Limits Evaluate the limit if it exists. $$\lim _{x \rightarrow-2} \frac{x^{2}-x+6}{x+2}$$

Step-by-Step Solution

Answer

The limit is -3.

Step 1: Identify the Indeterminate Form

First, substitute \( x = -2 \) into the function to check if it results in an indeterminate form. Substitute into the numerator and denominator separately: \(-2^2 - (-2) + 6 = 4 + 2 + 6 = 12\) and \(-2 + 2 = 0\). Since the denominator is zero, this is an indeterminate form \( \frac{12}{0} \).

Step 2: Factor and Simplify the Expression

Try to simplify the expression by factoring the numerator. Rewrite \( x^2 - x + 6 \) as \( x^2 - 4x + 3x + 6 \). By group factoring, it becomes \((x(x - 1) + 3(x - 1)) = (x - 1)(x + 2)\). Substitute back: \( \frac{(x - 1)(x + 2)}{x + 2} \).

Step 3: Cancel Common Factors

Cancel the common factor \((x + 2)\) from the numerator and denominator. The expression simplifies to \(x - 1\) for \(x eq -2\).

Step 4: Evaluate the Limit of the Simplified Function

Now evaluate the limit \( \lim_{x \to -2} (x - 1) \). Substitute \(x = -2\) into the simplified function: \(-2 - 1 = -3\).

Step 5: Conclusion of the Limit Evaluation

Since the simplified expression \(x - 1\) is continuous at \(x = -2\), the limit exists and equals \(-3\).

Key Concepts

Indeterminate Forms

When studying limits, one may encounter situations where substitution results in an expression that doesn't immediately provide an answer. These are known as indeterminate forms. They include scenarios like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), among others. The reason they are termed as such is because they do not help us directly conclude anything regarding the limit.

In the provided exercise, substituting \(x = -2\) into the function leads to a denominator of zero, resulting in an indeterminate form of \(\frac{12}{0}\). This signals that further analysis is required before determining the limit.

Handling indeterminate forms often involves simplifying the expression through factoring, rationalizing, or other algebraic manipulations. Understanding how to identify and approach these forms is essential for tackling limits in calculus effectively.

Factoring Expressions

Factoring is a crucial algebraic tool used to simplify functions, especially when dealing with limits. This technique can break down complex expressions into simpler products, making them more manageable to work with.

In our exercise, we aimed to simplify the expression \(x^2 - x + 6\). Through factoring by grouping, we rewrite the expression as \((x^2 - 4x) + (3x + 6)\). This can be further simplified to \((x(x - 1) + 3(x - 1)) = (x - 1)(x + 2)\).

Factoring plays a vital role in this problem because it reveals common factors in the numerator and the denominator. This step is necessary to effectively simplify rational expressions, eliminating indeterminate forms and easing the limit evaluation process.

Simplifying Rational Expressions

Once an expression is factored, the next step is to simplify it. Simplifying rational expressions involves canceling out common factors in the numerator and denominator. This is an important step, as it reduces the overall complexity and may resolve indeterminate forms.

In the original problem, after factoring, we obtain the expression \((x - 1)(x + 2)\) over \(x + 2\). We then cancel the common factor \(x + 2\), simplifying it to \(x - 1\).

This step is crucial since only through simplification can we evaluate the limit without the complications caused by a zero denominator. Always remember to identify and cancel out the common terms, but ensure that the values which made the denominator zero remain excluded.

Substitution in Limits

After an expression has been simplified, it often becomes easy to reevaluate the limit, this time without encountering indeterminate forms. Substitution is the final step where we substitute the value back into the simplified function to find the limit.

In our solution, the expression simplifies to \(x - 1\). By substituting \(x = -2\), we directly calculate \(-2 - 1 = -3\).

Substitution in limits is a straightforward process when simplifications are made correctly. It allows the evaluation of the function at points that originally caused indeterminate forms, thereby ensuring the correct calculation of the limit. It's this process that confirms the solution's accuracy.