Problem 68 - Calculus Single Variable

Question

For any real number \(k\) and any positive number \(a\) prove that $$ \lim _{x \rightarrow+\infty} \frac{x^{k}}{e^{a x}}=0 $$

Step-by-Step Solution

Answer

L'Hôpital's Rule shows that the polynomial grows slower than the exponential, leading the limit to zero.

Step 1: Initial Setup

To solve this limit problem, we need to consider two functions: a polynomial term, which is represented by \(x^k\) where \(k\) is any real number, and an exponential term, \(e^{ax}\) where \(a\) is positive. Our goal is to show that the polynomial term grows much slower than the exponential term as \(x\) approaches infinity.

Step 2: Applying L'Hôpital's Rule

The limit can initially be set up as an indeterminate form \(\frac{\infty}{\infty}\). In such cases, L'Hôpital's Rule can be applied which requires us to differentiate the numerator and the denominator until the limit can be evaluated. The rule states: \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}\) if the limit exists.

Step 3: Differentiate the Functions

Differentiate the numerator and denominator: the derivative of \(x^k\) is \(k \cdot x^{k-1}\), and the derivative of \(e^{ax}\) is \(a \cdot e^{ax}\). Thus, our expression becomes \(\lim_{x \rightarrow \infty} \frac{k \cdot x^{k-1}}{a \cdot e^{ax}}\).

Step 4: Reapply L'Hôpital's Rule

Notice that the new expression is also an indeterminate form \(\frac{\infty}{\infty}\). We can apply L'Hôpital's Rule again. Differentiating again gives us a new fraction: \(\lim_{x \rightarrow \infty} \frac{k(k-1) \cdot x^{k-2}}{a^2 \cdot e^{ax}}\).

Step 5: Continue Differentiation

We repeatedly apply L'Hôpital's Rule, each time reducing the exponent of the polynomial by 1, until the polynomial's derivative becomes zero when the power of \(x\) falls below zero. The exponential function, however, continues to have \(a^n \cdot e^{ax}\) in the denominator, which remains exponential and non-zero.

Step 6: Conclusion

Since the polynomial order reaches zero while the exponential function still grows rapidly, we find that subsequent derivatives ultimately lead to the numerator being zero and the limit turning to a form \(\frac{0}{e^{ax}}\), which equals zero as \(x\) approaches infinity.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that initially appear to be indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are called indeterminate because they don't have a clear value at first glance.

The rule helps us reevaluate such limits by focusing on the derivatives of the numerator and denominator. According to L'Hôpital's Rule, if you have a limit \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \) resulting in an indeterminate form, you can instead compute \( \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \), given this new limit exists.

Key points about L'Hôpital's Rule:

• It ONLY applies to forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Both functions \( f(x) \) and \( g(x) \) must be differentiable near \( c \).
• The limit of the derivatives must exist or reach a determined value.

By simplifying complex limits using derivatives, the rule provides us an easier path to finding their values.

Indeterminate Forms

Indeterminate forms are expressions that do not provide enough information about their value from a limit perspective. They commonly occur when evaluating the limits of certain functions. Forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) arise in problems involving asymptotic behaviors or undefined values.

Examples of indeterminate forms:

• \(\frac{0}{0}\) - Such as limits of functions approaching zero.
• \(\frac{\infty}{\infty}\) - Often occurs with exponential or polynomial growth.
• \