91Ó°ÊÓ

L'Hopital's Rule is a handy tool in calculus used for evaluating limits of indeterminate forms. It applies when you have a ratio of two functions where both the numerator and the denominator approach zero, or both approach infinity, as the variable approaches a specific value. To simplify and evaluate such limits,
follow these steps:

• Differentiate the numerator and the denominator separately.
• Take the limit of the new fraction obtained by these derivatives.

This rule simplifies complex limits and confirms results obtained by other methods.
In our problem, after applying L'Hopital's Rule to the original indeterminate form \( \frac{0}{0} \), the limit proved to be \( \frac{1}{4} \), corroborating the simplified expression.

Indeterminate forms occur when substituting a value into a mathematical expression yields results like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others. Such forms don't lead to a definite value directly, which makes it challenging to determine the limit.
In this exercise, substituting \( x = 2 \) into the original expression \( \frac{x-2}{x^2-4} \) resulted in \( \frac{0}{0} \), an indeterminate form.
When faced with indeterminate forms, alternative methods like factoring or L'Hopital's Rule are often used to resolve them and find the limit.

Factoring is a method used to simplify algebraic expressions. It involves expressing a polynomial as a product of its factors.
Factoring can often help in solving problems involving indeterminate forms by allowing us to cancel out and remove any zero-producing terms.
In the given exercise, the denominator \( x^2 - 4 \) is factored as \( (x-2)(x+2) \). This helps in simplifying the limit expression to \( \frac{1}{x+2} \) after canceling the common \( (x-2) \) term from both the numerator and the denominator. This makes the evaluation of the limit straightforward and direct.

Direct substitution is a straightforward method where you substitute the value of the variable into the function to find the limit. If the direct substitution gives a non-zero over non-zero value, the limit can be evaluated directly. However, if it results in indeterminate forms like \( \frac{0}{0} \) as in our problem, it indicates further work is needed.
In such cases, you'll usually need to try techniques like factoring, rationalizing, or applying L'Hopital's Rule. In this exercise, although direct substitution initially didn't work due to an indeterminate form, after simplifying by factoring, direct substitution was used again on the simpler expression \( \frac{1}{x+2} \), successfully yielding the result of \( \frac{1}{4} \).