91Ó°ÊÓ

L'Hôpital's Rule is a powerful tool for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these forms, the rule suggests taking derivatives of the numerator and the denominator separately, then trying the limit again. This process can often simplify complex expressions, making it easier to find the limit.
Although not directly used in this particular exercise, L'Hôpital's Rule is valuable to understand because the initial form of \( 1^{\infty} \) often needs transformation into a suitable form, like the ones mentioned earlier, for L'Hôpital's application.
Remember, it should only be applied if the function remains indeterminate after substitution, ensuring the correct and meaningful application of this method.

A natural logarithm transformation is particularly useful for simplifying expressions with powers, specifically those resulting in indeterminate forms like \( 1^{\infty} \).
In the given exercise, by setting \( y = \left(1-\frac{1}{x}\right)^{x} \), and then applying the natural logarithm, \( \ln(y) \) becomes \( x\ln\left(1-\frac{1}{x}\right) \).
This step transforms the complex power into a product involving a logarithm, which we can then tackle using well-known approximations for small values, making it much more manageable.
Logarithmic transformation is often a crucial step because it can linearize otherwise non-linear expressions, providing a bridge to easier mathematical manipulation and limit evaluation.

Evaluating limits often involves recognizing indeterminate forms and knowing how to handle them, whether by algebraic manipulation or applying special rules.
For this exercise, recognizing that \( \lim _{x \rightarrow \infty}\left(1-\frac{1}{x}\right)^{x} \) is an \( 1^{\infty} \) indeterminate form is key. It indicates that the expression will not straightforwardly yield a limit without transformation.
Once transformed through logarithms, we approach solving \( \lim_{x \to \infty} x\ln\left(1-\frac{1}{x}\right) \). By substituting the approximation \( \ln(1-u) \approx -u \) for small \( u \), the expression simplifies allowing us to solve for \( -1 \), aiding us to find that \( \lim_{x \to \infty} y = e^{-1} \).
Limit evaluation requires both creativity in manipulation and application of specific mathematical rules, as demonstrated here.

Exponential limits can often present complex challenges, especially when they form indeterminate expressions like \( 1^{\infty} \).
In this exercise, after evaluating the transformed expression and finding \( \lim_{x \to \infty} x\ln\left(1-\frac{1}{x}\right) = -1 \), we conclude with exponentiating both sides.
Exponentiation helps to revert the transformed expression back into an exponential form, giving us the actual limit sought—in this case, \( e^{-1} \) or \( \frac{1}{e} \).
Understanding how exponential functions behave in limits and knowing when to apply transformations like logarithms can simplify your work and lead directly to the solution.