91Ó°ÊÓ

When you attempt to evaluate limits in calculus, especially for rational functions, you may encounter results like \( \frac{0}{0} \). This is called an indeterminate form. It tells us that direct substitution didn’t work and we need a different strategy to find the limit.

Indeterminate forms often arise when both the numerator and the denominator of a fraction become zero as \( x \rightarrow c \). At this point, the limit is not immediately clear. Indeterminate forms indicate the need for algebraic manipulation to rework the expression into a form that can be evaluated conventionally.

To handle indeterminate forms, factorization is a powerful tool. Factorization involves rewriting a polynomial as a product of its factors, which can simplify expressions significantly. Let's break it down:

• Start by identifying common factors.
• Look for patterns such as the difference of squares or perfect square trinomials.
• Rearrange and simplify the expression by factoring completely.

In our example, the numerator \( x^2 + 2x - 3 \) can be factored as \( (x-1)(x+3) \), and the denominator \( 2x^2 + 3x - 9 \) factors to \( (2x-3)(x+3) \). This allows us to cancel common factors, which is crucial in simplifying the limit expression.

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions can have limits that result in indeterminate forms. Simplifying them by factorization often facilitates the evaluation of these limits.

Rational functions may have discontinuities at points where the denominator is zero because division by zero is undefined. In the exercise at hand, the expression originally had a zero at \( x = -3 \), which we resolved using algebraic simplification. This kind of precise manipulation helps in identifying limits more accurately.

Evaluating limits involves finding the value that a function approaches as the variable approaches a specific point. After simplifying the rational function, evaluate the limit by substituting the variable with the target value.

Post factorization and simplification, the indeterminate form was addressed by canceling out \( (x+3) \) from both the numerator and the denominator. This left us with \( \frac{x-1}{2x-3} \). When we substituted \(x = -3\) into this reduced form, we could easily calculate the limit as \( \frac{4}{9} \).

Key tips: ensure that any canceled terms just remove indeterminateness and do not introduce errors, and always verify the substitution for division by zero avoidance.