91Ó°ÊÓ

In calculus, limits help us understand the behavior of a function as it approaches a specific point. In our exercise, we are investigating what happens to the function \( f(x) = x e^{-x} \) as \( x \) tends towards positive infinity. These limits are crucial as they often simplify complex functions and provide insight into their long-term behavior.

When we talk about limits at infinity, like \( \lim_{x \to +\infty} \), we are often interested in the end behavior of a function. **Key Concepts to Remember:**

• Limits help determine asymptotic behavior, or how a function behaves as it approaches infinity or some finite value.
• Being familiar with exponential growth and decay is crucial, as exponential functions can grow or shrink very quickly.
• The concept of limits is foundational to calculate derivatives and integrals in calculus.

In our specific case, because the exponential decay in \( e^{-x} \) dominates the linear growth of \( x \), the product \( x e^{-x} \) approaches zero as \( x \) becomes infinitely large.

Indeterminate forms occur in calculus when evaluating a limit results in an ambiguous form due to infinity or zero interactions, such as \( \infty - \infty \) or \( 0 \cdot \infty \). The expression \( x e^{-x} \) at \( x \to +\infty \) initially gives us \( \infty \cdot 0 \), an indeterminate form that requires special techniques to evaluate.

**Strategies to Resolve Indeterminate Forms:**

• Rewriting the Expression: Convert the indeterminate form into a fraction, as was done with \( \frac{x}{e^x} \). This transformation often reveals opportunities for using calculus methods effectively.
• L'Hôpital's Rule: This rule is a powerful tool where both the numerator and denominator of a fraction are differentiated step by step, until a clear limit can be found.
• Examining Leading Terms: Consider the dominant behavior of functions, such as exponential terms that can significantly outweigh polynomial terms at infinity.

Applying L'Hôpital's Rule simplifies our indeterminate form \( \frac{\infty}{\infty} \) to a form easy to evaluate, leading us to the result that \( f(x) \to 0 \) as \( x \to +\infty \).

Exponential functions, such as \( e^x \), show a rapid growth or decay, and play a pivotal role in calculus due to their unique properties. In our problem, \( e^{-x} \) represents exponential decay, meaning it approaches zero as \( x \) increases.

**Understanding Exponential Functions:**

• Exponential Growth: Functions like \( e^x \) grow rapidly as \( x \) increases, making them larger than any polynomial function of \( x \).
• Exponential Decay: Conversely, \( e^{-x} \) shrinks towards zero rapidly, overpowering slower-growing terms such as linear ones.
• Calcualting Derivatives: The derivative of \( e^x \) remains \( e^x \), a unique feature making calculus with exponential functions straightforward and powerful.

In the exercise, the rapid decay of \( e^{-x} \) largely influences the behavior of \( f(x) = x e^{-x} \), pulling the entire expression towards zero as \( x \) approaches infinity. This influential quality of exponential decay underscores why understanding these functions is crucial in calculus.