91Ó°ÊÓ

In calculus, sometimes when computing limits, we encounter expressions that result in problematic forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. When you substitute a value directly into a rational expression and end up with such a form, you can’t determine the limit immediately. It signals a need for further algebraic manipulation resplendent with techniques like factoring or using L'Hôpital's Rule.

Recognizing an indeterminate form is crucial because it indicates that further simplification is necessary. In our given example, substituting \( x = -1 \) into \( \frac{x^2 + 6x + 5}{x^2 - 3x - 4} \) results in \( \frac{0}{0} \). This specific indeterminate form often suggests that there is a common factor in the numerator and denominator that can be canceled after factoring, which can resolve the indeterminacy.

Sometimes, if the expression is particularly complex, different mathematical approaches are necessary. Nonetheless, identifying these forms is the first step in effectively solving the problem.

Factoring is a method used in algebra to simplify expressions by breaking them down into a product of simpler terms. This technique is exceptionally useful in simplifying and evaluating limits of rational expressions. In our example, the quadratic expressions in both the numerator and the denominator are perfect candidates for factoring.

To factor the numerator, \( x^2 + 6x + 5 \), we look for two numbers that multiply to \( 5 \) (the constant term) and add to \( 6 \) (the linear coefficient). The numbers \( 5 \) and \( 1 \) satisfy these conditions, allowing us to write the numerator as \((x + 5)(x + 1)\).

Similarly, for the denominator \( x^2 - 3x - 4 \), we find numbers that multiply to \(-4\) and sum to \(-3\). The numbers \(-4\) and \(1\) satisfy these conditions, giving us \((x - 4)(x + 1)\).

Factoring simplifies many expressions, helping eliminate common factors, which is essential in solving problems involving indeterminate forms.

Once an expression is factored, the next step is simplifying it by canceling common factors in the numerator and denominator. This simplification is fundamental, especially in resolving indeterminate forms, as it transforms complex expressions into simpler ones more manageable for evaluation.

In the given problem, after factoring, we have the expression \( \frac{(x + 5)(x + 1)}{(x - 4)(x + 1)} \). The \( (x + 1) \) term appears in both the numerator and the denominator, making it a common factor that can be canceled out. This leaves us with the much simpler expression \( \frac{x + 5}{x - 4} \).

By reducing the expression, we remove the indeterminate form. After simplification, we can substitute \( x = -1 \) without issue. The limit becomes \( \frac{4}{-5} = -\frac{4}{5} \). Simplifying rational expressions is a powerful tool that makes it possible to calculate limits that might initially seem unsolvable.