Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(k\) be an arbitrary number and \(-1

Step-by-Step Solution

Answer

The limit is 0 as both terms tend to zero as \( n \to \infty \).

Step 1: Analyze the Expression

The given expression is \( \frac{k(k-1)(k-2) \cdots (k-n)}{n!} x^n \). We need to consider the behavior of this fraction as \( n \to \infty \). The denominator \( n! \) grows very rapidly as \( n \) increases, potentially faster than the numerator.

Step 2: Simplify the Expression with Stirling's Approximation

Using Stirling's approximation, we know \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \). The numerator is a product of \( n \) terms involving \( k \), which can be bounded by \( |k|^n \). Thus, our fraction approximately becomes \( \frac{|k|^n}{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n} x^n \).

Step 3: Consider the Exponential Term

We have \( \left( \frac{|k|}{\frac{n}{e}} \right)^n x^n \). Since \( -1 < x < 1 \), as \( n \) increases, \( x^n \to 0 \). The term \( \left( \frac{|k|}{n} \right)^n \) also tends toward zero because \( n \to \infty \) implies \( \frac{|k|}{n} < 1 \). Thus, this entire expression converges to zero, regardless of \( k \).

Step 4: Conclude the Proof

Both components of the fraction, \( x^n \) and \( \left( \frac{|k|}{n} \right)^n \), tend to zero as \( n \to \infty \). Therefore, the overall expression, \( \lim_{n \to \infty} \frac{k(k-1)(k-2) \cdots (k-n)}{n!} x^n = 0 \), holds true for any \( k \) and \( -1 < x < 1 \).

Key Concepts

Limits

In calculus, limits are vital for understanding the behavior of functions as they approach a particular point or infinity. When we look at the expression given in the exercise, we see that we are interested in what happens when "n" becomes infinitely large. It might help to think of limits as a way of "zooming out" to understand overall trends rather than focusing on specific values.

With the expression \( \frac{k(k-1)(k-2) \cdots (k-n)}{n !} x^n \), the focus is on how the function approaches zero as \( n \rightarrow \infty \), especially since \( -1 < x < 1 \). As \( n \) grows, \( x^n \to 0 \), because squaring a number between -1 and 1 always results in a smaller number unless it’s exactly zero or one.

There are a few essential points to remember when dealing with limits in calculus:

• Computing a limit involves determining how an expression behaves, not just how it is defined.
• Limits can often simplify complex expressions by focusing on ultimate behavior as variables grow without bounds.
• Recognizing that some terms diminish faster than others can be key to solving limit problems.

Stirling's Approximation

Stirling's approximation is a powerful tool in mathematics used to simplify factorials, especially useful when dealing with limits where \( n \) becomes very large. It gives us a way to approximate \( n! \) (n factorial) with a simpler expression. In our solution, we used the following form of Stirling's approximation:

• \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \)

This approximation tells us that factorials grow incredibly fast. For example, in our original exercise, \( n! \) in the denominator will soon outpace the terms in the numerator as \( n \rightarrow \infty \).

The numerator, noted as a product of successive terms involving \( k \), can be bounded by \( |k|^n \), showing its behavior as \( n \) increases. When paired with Stirling's approximation, this insight allows us to compare how the numerator and the rapidly growing denominator \( n! \) affect the whole fraction.

• By bounding the numerator under \( |k|^n \), we can simplify the limit question to something easier to digest.
• Stirling’s approximation clarifies that while both the numerator and denominator grow, \( n! \) grows faster.

Mathematical Proof

Mathematical proof is the backbone of verifying claims and understanding why mathematical statements are true. In our problem, a proof is needed to show the limit goes to zero. Here's the underpinning logic used in the given proof:

• The core of the problem hinges on analyzing how every part of the fraction behaves as \( n \rightarrow \infty \).
• Both components, the small \( x^n \) and \( \left( \frac{|k|}{n} \right)^n \), push the expression toward zero independently.

When proving statements with limits, especially ones involving infinite sequences or series, it is important to:

• Identify which parts of the expression dominate its behavior as \( n \) changes.
• Use approximations when necessary to simplify complex expressions, as with Stirling’s approximation for factorial growth.
• Ensure that all conditions (like \(-1 < x < 1\)) are appropriately respected throughout the proof.

By logical reasoning and breaking the problem into understandable parts, the proof strategically demonstrates that this fraction becomes insignificantly small, regardless of how large "k" might be.