Problem 26 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow \infty} x^{2} e^{-x} $$

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 \( x \rightarrow \infty \), \( x^2 \rightarrow \infty \) and \( e^{-x} \rightarrow 0 \) making the product an indeterminate form of type \( \infty \cdot 0 \). To apply l'Hospital's rule, we need a quotient, so we rewrite it as \( \cfrac{x^2}{e^x} \).

Step 2: Apply l'Hospital's Rule

Since the expression is now \( \frac{x^2}{e^x} \), the indeterminate form as \( x \rightarrow \infty \) becomes \( \frac{\infty}{\infty} \). This allows us to apply l'Hospital's rule by differentiating the numerator and denominator.Compute the derivatives: The derivative of \( x^2 \) is \( 2x \), and the derivative of \( e^x \) is \( e^x \).

Step 3: Evaluate the New Limit

Apply l'Hospital's rule once, giving the new limit:\[ \lim _{x \rightarrow \infty} \frac{2x}{e^x} \]This is again an \( \frac{\infty}{\infty} \) form, so apply l'Hospital's rule a second time, giving:Compute the derivatives again: The derivative of \( 2x \) is \( 2 \), and the derivative of \( e^x \) remains as \( e^x \).

Step 4: Final Evaluation using l'Hospital's Rule Again

After applying l'Hospital's rule a second time, evaluate the new limit:\[ \lim _{x \rightarrow \infty} \frac{2}{e^x} \]As \( x \rightarrow \infty \), \( e^x \rightarrow \infty \), so this becomes \( \frac{2}{\infty} = 0 \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, particularly those that involve functions increasing or decreasing without bound, we often encounter indeterminate forms. These are expressions where the limit seems difficult to deduce directly. In the example of \( \lim_{x \rightarrow \infty} x^2 e^{-x} \), we find ourselves facing the form \( \infty \times 0 \). This form occurs because while \( x^2 \) grows infinitely large, \( e^{-x} \) shrinks toward zero. Such situations are problematic because multiplying infinity by zero does not lead to a clear numerical result. They require more sophisticated mathematical tools, like l'Hospital's Rule, to resolve. In general, indeterminate forms include:

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

Understanding these forms is essential for solving complex limits in calculus, as recognizing them allows for the appropriate use of techniques like l'Hospital's Rule.

Exponential Functions

Exponential functions are a key part of calculus and frequently appear in limit problems. These functions, outlined as \( f(x) = a^x \), where \( a \) is a constant, exhibit rapid growth or decay. In our specific problem, the function \( e^{-x} \) plays a crucial role. Here, \( e \) is the base of the natural logarithm, approximately 2.718. When raised to the power of a negative \( x \), the function starts at 1 when \( x = 0 \) and descends towards 0 as \( x \) increases. Some important properties of exponential functions include:

• Continual growth (if the exponent is positive) or decay (if the exponent is negative).
• The derivative of \( e^x \) is \( e^x \), and the derivative of \( e^{-x} \) is \( -e^{-x} \).
• An exponential function grows or decays proportionally at a constant rate.

Understanding exponential functions helps in predicting behaviors in natural and physical sciences, and they often align with real-world data in growth and decay scenarios.

Calculus Limit Problems

Calculus limit problems frequently require deep understanding and thoughtful approaches. They involve predicting the behavior of functions as they approach a certain point or tend towards infinity.The primary goal with limit problems is to understand how a function behaves near boundaries or at infinity, which can apply to various fields such as physics, engineering, and economics. Limit problems such as \( \lim_{x \rightarrow \infty} x^2 e^{-x} \) represent a common scenario where advanced techniques might be necessary. When you encounter a limit that leads to an indeterminate form, it's a signal to potentially utilize l'Hospital's Rule if the setup involves a quotient.Steps to solve these problems more effectively:

• Identify the type of limit and potential indeterminate form that appears.
• Check if l'Hospital's Rule can be used, typically for \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiate the numerator and denominator separately, and compute the new limit.
• Repeat the differentiation if the indeterminate form persists after the first application of l'Hospital's Rule.

Mastering calculus limit problems gives insights into instantaneous change, rates, and values at boundaries, forming a cornerstone of differential calculus understanding.