Question

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

Short answer

The limit is 0.

Step-by-step solution

Identify the Form

First, we need to recognize the mathematical form of the given limit as \( x \to \infty \). The given function is \( x^2 e^{-x} \), which can be written as \( \frac{x^2}{e^x} \). As \( x \to \infty \), \( x^2 \to \infty \) and \( e^x \to \infty \). This results in an \( \frac{\infty}{\infty} \) indeterminate form.

Apply L'Hôpital's Rule

Since the form is \( \frac{\infty}{\infty} \), we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator. First derivative of the numerator, \( x^2 \), is \( 2x \). The derivative of the denominator, \( e^x \), is \( e^x \). So the limit becomes: \[\lim_{x \to \infty} \frac{2x}{e^x}\]

Apply L'Hôpital's Rule Again

The new expression \( \frac{2x}{e^x} \) is still in the \( \frac{\infty}{\infty} \) form, so we apply L'Hôpital's Rule again. Now, differentiate the numerator \( 2x \) to get \( 2 \); differentiate the denominator \( e^x \) to get \( e^x \). The limit now is:\[\lim_{x \to \infty} \frac{2}{e^x}\]

Evaluate the Final Limit

As \( x \to \infty \), \( e^x \to \infty \) and hence \( \frac{2}{e^x} \to 0 \). Therefore, the value of the original limit is 0.

Key concepts

Indeterminate Forms

When evaluating limits, you may encounter expressions that don't lead straightforwardly to a numerical conclusion. These expressions are known as indeterminate forms. A common example, as in the exercise provided, is the form \( \frac{\infty}{\infty} \). In essence, indeterminate forms occur when substituting the limit value into the function results in an ambiguous expression.
This can arise with limiting processes involving infinity, zero, or both. For example, other notable indeterminate forms include \( 0/0 \), \( 0\times\infty \), or \( \infty - \infty \). Indeterminate forms do not imply that a limit does not exist but rather that the form itself cannot directly determine the limit. Therefore, additional techniques, like algebraic manipulation or L'Hôpital's Rule, are often necessary to evaluate these limits accurately.

Limit Evaluation

Determining the value of a limit involves several techniques, one of which may include L'Hôpital's Rule, as demonstrated in the solution to the problem. The essence of limit evaluation is to find out what value a function approaches as the input approaches a specific point.
When direct substitution into a function yields an indeterminate form, such as \( \frac{\infty}{\infty} \) or \( 0/0 \), other methods must be employed. These methods may involve:

• Algebraic manipulation to simplify the function before substituting the limit.
• L'Hôpital's Rule, which involves differentiating the numerator and the denominator to resolve the indeterminate form.
• Series expansion or trigonometric identities for more complex expressions.

The goal is to transform the expression into a form that permits a clear evaluation. In our exercise, after identifying the indeterminate form, L'Hôpital's Rule enabled us to compute the limit effectively.

Differentiation

Differentiation is a fundamental concept in calculus employed to determine the rate at which a function changes at any given point. In the context of L'Hôpital's Rule, it plays a crucial role in resolving indeterminate forms like \( \frac{\infty}{\infty} \).
When faced with an indeterminate form, L'Hôpital's Rule suggests differentiating the numerator and the denominator separately, which often leads to simpler functions that can be evaluated more directly. In the provided example:

• The function \( f(x) = x^2 \) was differentiated to yield \( 2x \).
• The function \( g(x) = e^x \) was differentiated to yield \( e^x \) again, as it is its own derivative.
• This led us to a simpler analysis of \( \frac{2}{e^x} \), which could be directly evaluated as \( x \to \infty \).

Through differentiation, complex limits become approachable, offering a pathway to a clear solution.