Problem 20 - Thomas Calculus

Question

The function \(e^{x}\) outgrows any polynomial Show that \(e^{x}\) grows faster as \(x \rightarrow \infty\) than any polynomial $$ a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} $$

Step-by-Step Solution

Answer

Exponential growth outpaces polynomial growth as \(x \to \infty\).

Step 1: Express the Growth Rates

To show that \(e^{x}\) grows faster than any polynomial \(p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\), we need to compare the growth rates. As \(x \rightarrow \infty\), the dominant term of the polynomial is \(a_n x^n\). Therefore, the growth rate of the polynomial can be approximated by its highest degree term \(x^n\).

Step 2: Use L'Hôpital's Rule

Consider the limit \(\lim_{x \to \infty} \frac{e^x}{x^n}\). By using L'Hôpital's Rule, which is applicable here as both numerator and denominator approach infinity, derive the limit by differentiating the numerator and the denominator until the exponential term dominates the comparison.

Step 3: Apply L'Hôpital's Rule Iteratively

Repeat applying L'Hôpital's Rule. The differentiation of the numerator \(e^x\) results in \(e^x\), while the differentiation of \(x^n\) is \(nx^{n-1}\). Iteratively apply the rule (a total of \(n\) times) until the polynomial's term becomes a constant while \(e^x\) continues as \(e^x\).

Step 4: Calculate the Limit

After applying L'Hôpital's Rule \(n\) times, the limit simplifies to \(\lim_{x \to \infty} \frac{e^x}{n!} = \infty\). This shows that as \(x\) approaches infinity, \(e^x\) grows without bound, whereas any \(x^n\) results in a finite value. Hence \(e^x\) overtakes \(x^n\), proving its dominant growth rate.

Key Concepts

Polynomials

Polynomials are mathematical expressions consisting of variables and coefficients. They are constructed from basic operations such as addition, subtraction, and multiplication. A general polynomial can be expressed as:

• a sum of terms, each being a constant multiplied by a variable raised to a non-negative integer power
• an example of a polynomial expression could be: \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)

Where:

• \( a_n, a_{n-1}, \ldots, a_0 \) are the coefficients
• \( x^n, x^{n-1}, \ldots, x^0 \) are the terms with \( x^n \) being called the dominant term for large \( x \)

A polynomial's degree is determined by the highest power of its variable term. In the large \( x \) behavior of a polynomial, the highest power term dictates its growth rate as it overshadows all lower power terms. Therefore, when considering the behavior of polynomials as \( x \to \infty \), the term \( a_n x^n \) is primarily used as it dominates the expression.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used to evaluate limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these scenarios, L'Hôpital's Rule provides a method to simplify complex limits:

• Take the derivative of the numerator and the derivative of the denominator separately.
• Evaluate the limit of the resulting expression.

For example, consider the limit \( \lim_{x \to \infty} \frac{e^x}{x^n} \):

• Both \( e^x \) and \( x^n \) head towards infinity, qualifying the use of L'Hôpital's Rule.
• Differentiate \( e^x \) to get \( e^x \) and differentiate \( x^n \) to get \( n x^{n-1} \).
• Repeat the differentiation \( n \) times essentially reducing \( x^n \) to a constant.

This process ultimately reveals that the exponential \( e^x \) grows at a faster rate than the polynomial, as the resulting expression heads towards infinity.

Infinity Limits

Infinity limits explore the behavior of functions as the input grows without bound, either positively or negatively. Specifically, examining infinity limits involves:

• Understanding which parts of an expression dominate as the variable approaches infinity.
• Using methods like L'Hôpital's Rule to resolve complex relationships.

When studying how functions behave as \( x \to \infty \):

• We often compare terms to identify the dominant growth rate.
• This is particularly important when comparing exponential functions like \( e^x \) with polynomials.

In our context, as \( x \to \infty \), the exponential \( e^x \) and polynomials behave differently:

• Polynomials grow at a rate proportional to their highest degree term, \( x^n \).
• Exponential functions such as \( e^x \) grow by multiplying by a constant base, implying they increase much faster than any polynomial.

Understanding these behaviors is crucial in calculus as it helps in predicting and analyzing the long-term behavior of mathematical models.