Problem 387 - Calculus Volume 1

Question

For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow \infty} x^{2} e^{-x} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

The given limit is \[\lim _{x \rightarrow \infty} x^{2} e^{-x}\] As \(x\) approaches \(\infty\), \(x^2\) grows towards infinity and \(e^{-x}\) approaches 0. This is an indeterminate form of type \(\infty \times 0\). To use L'Hôpital's Rule, we need to convert this into a quotient form that results in either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Step 2: Rewrite the Expression

Rewrite \(x^2 e^{-x}\) as a quotient.\[x^2 e^{-x} = \frac{x^2}{e^x}\]Now we have the expression \(\frac{x^2}{e^x}\) which forms an \(\frac{\infty}{\infty}\) type indeterminate form as \(x\) approaches \(\infty\).

Step 3: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule, which states that for \(\frac{\infty}{\infty}\) forms, the limit of the quotient of the derivatives can be taken: \[\lim _{x \rightarrow \infty} \frac{x^2}{e^x}\]Take the derivative of the numerator and the denominator:

Step 4: Differentiate the Numerator and Denominator

The derivative of \(x^2\) is \(2x\).The derivative of \(e^x\) is \(e^x\).Apply L'Hôpital's Rule after taking derivatives:\[\lim _{x \rightarrow \infty} \frac{2x}{e^x}\] This still forms an \(\frac{\infty}{\infty}\) type indeterminate form.

Step 5: Apply L'Hôpital's Rule Again

Since the limit \(\frac{2x}{e^x}\) is still an indeterminate form of \(\frac{\infty}{\infty}\), apply L'Hôpital's Rule again by differentiating the numerator and denominator again.

Step 6: Differentiate Again

Differentiate the numerator \(2x\) to get \(2\).Differentiate the denominator \(e^x\) to get \(e^x\).Apply L'Hôpital's Rule again:\[\lim _{x \rightarrow \infty} \frac{2}{e^x}\]

Step 7: Evaluate the Limit

As \(x\) approaches infinity, \(e^x\) approaches infinity, so:\[\frac{2}{e^x} \rightarrow 0\]Thus, the limit evaluates to \(0\).

Key Concepts

Calculus Limits

In calculus, limits help us analyze the behavior of functions as they approach specific points or infinity. They are the foundation for understanding how functions behave near boundaries. For instance, in the given exercise, we need to evaluate the limit of the expression \(\lim _{x \rightarrow \infty} x^2 e^{-x}\). This means we want to see what happens to this expression as \(x\) becomes infinitely large. Limits are crucial in calculus because they lead to the core ideas of continuity, derivatives, and integrals.

• They help us understand the behavior of functions at points where they may not be explicitly defined.
• They're essential for the formulation of derivatives, telling us how a function changes at any given point.
• Limits are key in calculating eventual behavior as input values grow exponentially large or small.

Indeterminate Forms

Indeterminate forms represent expressions that do not initially indicate a definitive limit as variables get larger, smaller, or approach certain values. In this exercise, we encounter an indeterminate form of \(\infty \times 0\), which occurs frequently in calculus. Indeterminate forms include:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \times \infty\)
• \infty - \infty\
• 1^{\infty}\

These forms require special techniques like L'Hôpital's Rule, algebraic manipulation, or series expansions to resolve. Recognizing an expression as an indeterminate form is the first step in applying the appropriate mathematical tools to find limits.

Derivatives Application

Derivatives play a pivotal role in using L'Hôpital’s Rule to find limits, especially when indeterminate forms occur. Derivatives tell us the rate at which a function changes and are used here to simplify complex expressions to determine their limits.
In the example, L'Hôpital's Rule is applied to the quotient \(\frac{x^2}{e^x}\), transforming our indeterminate \(\infty/\infty\) form into something manageable, \(\frac{2x}{e^x}\) and then \(\frac{2}{e^x}\) after differentiating again:

• First derivative of numerator \(: \frac{d}{dx}(x^2) = 2x\)
• First derivative of denominator \(: \frac{d}{dx}(e^x) = e^x\)
• Second derivative of numerator \(: \frac{d}{dx}(2x) = 2\)
• Second derivative of denominator \(: \frac{d}{dx}(e^x) = e^x\)

Without derivatives, evaluating these limits for certain types of functions would be far more complex.

Infinity in Calculus

Infinity plays a special role in calculus, allowing us to explore the behavior of functions as they move beyond finite bounds. When evaluating limits towards infinity, it's vital to understand how functions behave asymptotically.
In the given problem, the function \(x^2\) tends to infinity, and \(e^{-x}\) shrinks towards zero, creating a balancing act that leads us to use L'Hôpital’s Rule. This helps us make sense of the expression \(\lim _{x \rightarrow \infty} x^2 e^{-x}\) as \(x\) becomes very large.
Calculating limits involving infinity lets us verify how functions might behave at extreme values, offering insights like:

• Infinite growth or decay scenarios
• Calculating horizontal asymptotes
• Understanding long-term trends in function behaviors

With infinite limits, we cover extremes that are crucial in both theoretical and applied mathematics.