91Ó°ÊÓ

In calculus, simplifying rational expressions is a crucial step before evaluating limits. A rational expression is a fraction where the numerator and the denominator are polynomials. Consider the given expression: \[ \frac{x+2}{x^{2}-4} \]The denominator, \( x^{2}-4 \), is a difference of squares, which can be factored as:

• \( x^{2} - 4 = (x+2)(x-2) \)

Once factored, the expression becomes:\[ \frac{x+2}{(x+2)(x-2)} \]At this point, we notice that \( x+2 \) in the numerator and denominator can cancel each other out, but only for values where \( x eq -2 \). The simplified form is:\[ \frac{1}{x-2} \]It's important to maintain the condition that \( x eq -2 \) since dividing by zero is undefined. Simplifying expressions can make limits easier to evaluate, especially near points where initial expressions seem complex.

After simplifying the expression, the next step in calculus is to evaluate the limit. Let's follow through using our simplified expression:\[ \lim_{x \rightarrow -2} \frac{1}{x-2} \]To find this limit, we substitute \( x = -2 \) into the expression:\[ \frac{1}{-2-2} = \frac{1}{-4} \]This calculation gives us the limit value. Evaluating limits helps determine the behavior of functions as they approach a particular point. In this scenario, even though the function seemed undefined at \( x = -2 \) initially due to the cancellation in simplification, the limit exists and is finite at \( \frac{1}{-4} \).
Always remember to consider the domain and check for simplification steps that might be affecting it.

In calculus, indeterminate forms often arise when evaluating limits. These are expressions where direct substitution initially leads to an undefined form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the given problem, we first had:\[ \frac{x+2}{x^{2}-4} \]When \( x \rightarrow -2 \), both the numerator \( x+2 \) and the factor \( x+2 \) in the denominator become zero, leading to an indeterminate form \( \frac{0}{0} \). This situation invites a deeper analytic approach, such as:

• Simplifying by factoring and reducing the expression when possible.
• Using L'Hôpital's Rule if applicable, which involves differentiating the numerator and denominator separately and then taking the limit.

For the exercise given, simplifying and canceling terms successfully resolved the indeterminate form, enabling a direct evaluation of the limit.