Problem 64 - Calculus for Biology and Medicine

Question

Show that $$ \lim _{x \rightarrow \infty} x^{p} e^{-x}=0 $$ for any positive number \(p .\) Graph \(f(x)=x^{p} e^{-x}, x>0\), for \(p=1 / 2,1\), and \(2 .\) Since \(f(x)=x^{p} e^{-x}=x^{p} / e^{x}\), the limiting behavior \(\left(\lim _{x \rightarrow \infty} \frac{x^{p}}{e^{x}}=0\right)\) shows that the exponential function grows faster than any power of \(x\) as \(x \rightarrow \infty\).

Step-by-Step Solution

Answer

As \( x \to \infty \), \( e^x \) grows faster than any \( x^p \), showing that \( \lim_{x \to \infty} x^p e^{-x} = 0 \).

Step 1: Analyze the Given Function

The function given is \( f(x) = x^p e^{-x} = \frac{x^p}{e^x} \). To show that \( \lim_{x \to \infty} x^p e^{-x} = 0 \), we note that the denominator \( e^x \) grows exponentially while the numerator \( x^p \) grows polynomially. Exponential growth dominates polynomial growth as \( x \to \infty \).

Step 2: Use L'Hôpital's Rule

Since both numerator and denominator approach infinity as \( x \to \infty \), we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided the limit on the right also exists.

Step 3: Differentiate the Numerator and Denominator

Differentiate the numerator and the denominator: - \( f(x) = x^p \), so \( f'(x) = p x^{p-1} \).- \( g(x) = e^x \), so \( g'(x) = e^x \).Then, applying L'Hôpital's Rule gives \( \lim_{x \to \infty} \frac{x^p}{e^x} = \lim_{x \to \infty} \frac{p x^{p-1}}{e^x} \).

Step 4: Repeat L'Hôpital's Rule if Necessary

For \( \lim_{x \to \infty} \frac{p x^{p-1}}{e^x} \), the numerator grows polynomially, and the denominator grows exponentially. If \( x^{p-1} \) is still larger than the exponential \( e^x \), repeat L'Hôpital's Rule by further differentiation until the numerator becomes constant or zero, while the denominator \( e^x \) remains exponential.

Step 5: Evaluate the Limit

After repeated application of L'Hôpital's Rule \( p \) times, the derivative of the numerator will eventually become zero, when the initial \( x^p \) term reduces to a constant or zero. Thus, \( \lim_{x \to \infty} x^p e^{-x} = 0 \), as the exponential term in the denominator will always grow faster than any power of \( x \).

Step 6: Graph the Function for Different Values of \( p \)

For graphical illustration, plot \( f(x) = x^{p} e^{-x} \) for \( p = \frac{1}{2}, 1, \) and \( 2 \). Observe that as \( x \to \infty \), the function approaches zero faster for larger values of \( p \) because the exponential decay dominates the polynomial growth more strongly.

Key Concepts

Exponential Functions

Exponential functions appear often in mathematics due to their unique property of having extremely fast growth rates. They are defined by the expression \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. This function is characterized by its constant relative growth rate: the derivative of an exponential function is proportional to the function itself, \( \frac{d}{dx} e^x = e^x \). This means that as \( x \) increases, so does the rate at which \( e^x \) increases.
Exponential functions are used to model phenomena that grow rapidly, such as populations or investments with continuous compounding interest. Because of their exponential nature, they outgrow polynomial functions—which increase at a slower, polynomial rate—by a significant margin as \( x \) approaches infinity. This difference is the key to understanding why an exponential denominator, \( e^x \), in a fraction like \( \frac{x^p}{e^x} \) will dominate any polynomial growth in the numerator, leading to the limit of zero as \( x \to \infty \).
In summary, the exponential function's rapid growth rate allows it to overpower polynomial expressions, making it crucial in understanding the behavior of functions and limits involving powers of \( x \) divided by \( e^x \).

Polynomial Growth

Polynomial growth refers to the rate at which functions built from sums of powers of \( x \) increase. These functions are commonly expressed as \( x^p \), where \( p \) is a positive number. As \( x \) grows, so does \( x^p \), but at a much slower rate compared to exponential functions like \( e^x \).
Polynomials can have various growth patterns, dependent on the degree of \( p \). For example:

• Linear growth occurs when \( p = 1 \), which is relatively slow.
• Quadratic growth happens with \( p = 2 \), increasing a bit faster.
• Cubic and higher-degree polynomials continue this trend.

However, despite their different rates of increase, all polynomial functions share the characteristic of having a finite derivative at each point. This means that the rate of change is not constant, unlike exponential functions.
In the context of finding limits like \( \lim_{x \to \infty} \frac{x^p}{e^x} = 0 \), polynomial functions represent the "slow grower," resulting in the overall fraction approaching zero as the exponential denominator increases at a much faster rate. This deep understanding of polynomial behavior is essential when performing limit calculations and comparisons with exponential growth.

L'Hôpital's Rule

When evaluating limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule is a valuable tool. This rule provides a method to simplify these expressions by differentiating the numerator and the denominator separately.

• If \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
  \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \)

L'Hôpital’s Rule simplifies the problem by taking successive derivatives of the numerator and denominator until a determinate form is reached.
In the given exercise \( \frac{x^p}{e^x} \), applying L'Hôpital's Rule involves differentiating:
- The numerator \( x^p \) differentiates to \( p x^{p-1} \), decreasing its power.
- The denominator \( e^x \) remains as \( e^x \).
Repeating this process reduces the numerator's power continuously until achieving a constant, eventually reaching a limit where the numerator is essentially zero while the denominator is exponentially large. This establishes the final result that \( \lim_{x \to \infty} \frac{x^p}{e^x} = 0 \). Such insights provide clarity when analyzing limits and tackling complex calculus problems.