Problem 59 - Calculus

Question

This exercise shows that \(F(x)=x^{n} / e^{x} \rightarrow 0\) as \(x \rightarrow \infty\) (a) Find \(d F / d x\). Notice that \(F(x)\) decreases for \(x>n>0\) The maximum of \(x^{n} / e^{x},\) at \(x=n,\) is \(n^{n} / e^{n}\) (b) \(F(2 x)=(2 x)^{n} / e^{2 x}=2^{n} x^{n} / e^{x} \cdot e^{x} \leqslant 2^{n} n^{0} / e^{n} \cdot e^{x}\) Deduce that \(F(2 x) \rightarrow 0\) as \(x \rightarrow \dot{\infty}\). Thus \(F(x) \rightarrow 0\)

Step-by-Step Solution

Answer

\(F(x) \rightarrow 0\) as \(x \rightarrow \infty\).

Step 1: Differentiate F(x)

To find \(\frac{dF}{dx}\) where \(F(x) = \frac{x^n}{e^x}\), we apply the quotient rule. The quotient rule states that \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\). Here, \(u = x^n\) and \(v = e^x\). We calculate \(\frac{du}{dx} = nx^{n-1}\) and \(\frac{dv}{dx} = e^x\). Therefore, \(\frac{dF}{dx} = \frac{e^x \cdot nx^{n-1} - x^n \cdot e^x}{e^{2x}} = \frac{e^x(nx^{n-1} - x^n)}{e^{2x}} = \frac{nx^{n-1} - x^n}{e^x}\).

Step 2: Analyze the behavior for x > n > 0

For \(x > n > 0\), notice that \(nx^{n-1} < x^n\) because the term \(x^{n-1}\) multiplied with n eventually becomes smaller than \(x^n\) as \(x\) grows. Thus, the numerator \(nx^{n-1} - x^n\) becomes negative, indicating that \(F(x)\) is decreasing for \(x > n\). The maximum occurs at \(x = n\), where \(F(n) = \frac{n^n}{e^n}\).

Step 3: Analyze F(2x)

Consider \(F(2x) = \frac{(2x)^n}{e^{2x}} = \frac{2^n x^n}{e^{x} \cdot e^x}\). This can be rewritten as \(\frac{2^n x^n}{e^{x}} \times \frac{1}{e^x}\). If \(x \geq n\), then \(x^n \leq n^n\), so \(\frac{2^n x^n}{e^x} \leq \frac{2^n n^n}{e^x}\). As \(x\) increases, \(e^x\) grows very fast, making the fraction \(\frac{2^n n^n}{e^x}\) go to 0.

Step 4: Conclude for F(x) as x approaches infinity

Since \(F(2x) \rightarrow 0\) as \(x \rightarrow \infty\), and \(F(x)\) is decreasing for \(x > n\), it follows that \(F(x) \rightarrow 0\) as \(x \rightarrow \infty\) as well.

Key Concepts

Function Limits

Understanding the concept of limits is crucial in calculus, especially when analyzing the behavior of functions as they approach certain values. In this exercise, we are interested in how the function \( F(x) = \frac{x^n}{e^x} \) behaves as \( x \rightarrow \infty \). The limit helps us determine the behavior of the function as the input value becomes extremely large. To find the limit of \( F(x) \) as \( x \rightarrow \infty \), observe that the exponential function \( e^x \) in the denominator grows much faster than the polynomial function \( x^n \) in the numerator. This means that as \( x \) gets larger, the overall value of \( F(x) \) becomes smaller, tending towards zero. Thus, mathematically, we express this by stating that \( \lim_{{x \to \infty}} \frac{x^n}{e^x} = 0 \). This simplifies the analysis of the function's behavior and helps us conclude that \( F(x) \) approaches zero as \( x \) approaches infinity.

Quotient Rule

The quotient rule is a fundamental tool in calculus for differentiating functions that are written as one function divided by another. It is particularly useful in this exercise for differentiating the function \( F(x) = \frac{x^n}{e^x} \).According to the quotient rule, if you have a function \( \frac{u}{v} \), its derivative is given by: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]For \( F(x) \), we let \( u = x^n \) and \( v = e^x \). After computing, \( \frac{du}{dx} = nx^{n-1} \) and \( \frac{dv}{dx} = e^x \).Substituting these into the quotient rule formula, we find:\[ \frac{dF}{dx} = \frac{e^x(nx^{n-1}) - x^n(e^x)}{(e^x)^2} = \frac{nx^{n-1} - x^n}{e^x} \]This result helps in understanding how the function's slope or gradient changes across its domain, which is essential for analyzing its behavior.

Exponential Functions

Exponential functions, particularly \( e^x \), play a significant role in this exercise. They exhibit rapid growth, which impacts the behavior of the function \( F(x) = \frac{x^n}{e^x} \).The exponential function \( e^x \) grows faster than any power of \( x \) as \( x \rightarrow \infty \). This fast growth is why \( \frac{x^n}{e^x} \) tends to zero as \( x \) increases. The dominance of the exponential term influences the overall behavior of the quotient, causing it to decrease towards zero as \( x \rightarrow \infty \).Understanding this characteristic is crucial, especially when considering functions involving exponential terms. It provides insight into the asymptotic behavior of functions, allowing for predictions about their long-term trends.

Behavior Analysis of Functions

Behavior analysis of functions involves understanding how a function changes, whether it increases, decreases, or achieves a maximum or minimum value. For the function \( F(x) = \frac{x^n}{e^x} \), behavior analysis gives us insights into its trends over its domain.For \( x > n \), we can see that \( F(x) \) is decreasing because the term \( nx^{n-1} \) in the numerator of the derivative \( \frac{nx^{n-1} - x^n}{e^x} \) becomes smaller than \( x^n \). This leads to a negative derivative, indicating a decline in function values beyond \( x=n \). The maximum value occurs at \( x=n \) where \( F(n) = \frac{n^n}{e^n} \). This point marks the peak after which the function decreases. By examining \( F(2x) \) and other transformations, we can further substantiate the decreasing trend as \( x \rightarrow \infty \). Through this analysis, we conclude that \( F(x) \rightarrow 0 \) as \( x \rightarrow \infty \), confirming the function's behavior and understanding the impact of exponential growth on polynomial terms.