91Ó°ÊÓ

L'Hopital's Rule is a powerful tool in calculus that helps evaluate limits of indeterminate forms like \(\frac{\text{0}}{\text{0}}\) or \(\frac{\text{∞}}{\text{∞}}\). In our problem, we encounter the indeterminate form \(\frac{x^{2}}{e^{k x}}\) as \(x \to \text{∞}\). Instead of directly trying to solve the limit, we can apply L'Hopital's Rule by taking derivatives. By repeatedly differentiating the numerator and the denominator, we eventually simplify the limit to a form where it can be easily evaluated. In our case:
\(\frac{x^{2}}{e^{k x}} \to \frac{2x}{k e^{k x}} \to \frac{2}{k^{2} e^{k x}}\)
Each differentiation brings the limit closer to a solvable form. Eventually, we show that the original function approaches zero as \(x \to \text{∞}\).

Exponential decay is a process where quantities decrease rapidly over time. In our function \(e^{-k x}\), the exponential part represents exponential decay because \(-k x\) means that the exponent is negative, making the function decay faster as \(x\) increases. When \(-kx\) becomes very large negatively, \(e^{-k x}\) approaches zero extremely quickly.
The key idea is that exponential decay surpasses polynomial growth in terms of how fast it approaches zero. This is why in \(x^{2} e^{-k x}\), the exponential decay dominates and pulls the entire expression towards zero, no matter how large the polynomial term \(x^2\) becomes.

Polynomial growth refers to functions where the variable appears raised to some power, such as \(x^{2}\) in our example. Polynomial functions grow without bound as the variable increases. This means the term \(x^2\) becomes very large as \(x\) increases. However, polynomial growth is slower compared to exponential functions, particularly exponential decay.
Even though \(x^2\) keeps growing indefinitely, in the context of our problem, its growth isn't fast enough to counteract the rapid decrease of \(e^{-k x}\). This is a crucial point in understanding why the whole function \(x^{2} e^{-k x}\) trends towards zero rather than infinity or another value.

The concept of a limit at infinity helps us understand the behavior of a function as the input grows larger and larger. Specifically, we are interested in what value the function approaches.
For the function \(x^{2} e^{-k x}\), we are analyzing its limit as \(x \to \text{∞}\). By using L'Hopital's Rule, we can rigorously show that this limit is zero. Without using calculus tools, it may not be obvious; however, applying the rule provides a structured way to follow the growth and decay dynamics of each part of the function.
Our step-by-step simplification ultimately demonstrates that \(x^{2} e^{-k x} \to 0\) as \(x \to \text{∞}\).