Problem 121 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty}\left(\sin \frac{\pi}{2 n} \cdot \sin \frac{2 \pi}{2 n} \cdot \sin \frac{3 \pi}{2 n} \cdots \sin \frac{(n-1) \pi}{n}\right)^{1 / n}\) (A) \(\frac{1}{4}\) (B) 4 (C) 1 (D) None of these

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{1}{2}\); therefore, the answer is (D) None of these.

Step 1: Understand the Problem

We are given a limit problem where we need to find the value as \( n \) approaches infinity for the expression \( \left( \sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n} \cdot \sin \frac{3\pi}{2n} \cdots \sin \frac{(n-1)\pi}{n} \right)^{1/n} \). Our goal is to determine the correct value among the given options.

Step 2: Expression Simplification

First note that the expression inside the product is \( \sin \frac{k\pi}{2n} \) for \( k = 1, 2, 3, \ldots, n-1 \). As \( n \to \infty \), each \( \sin \frac{k\pi}{2n} \approx \frac{k\pi}{2n} \) because for small values of \( x \), \( \sin x \approx x \).

Step 3: Analyze the Product

The expression simplifies to a product: \( \prod_{k=1}^{n-1} \frac{k\pi}{2n} = \left( \frac{\pi}{2n} \right)^{n-1} \prod_{k=1}^{n-1} k = \left( \frac{\pi}{2n} \right)^{n-1} (n-1)! \).

Step 4: Use Stirling’s Approximation

Apply Stirling’s approximation to the factorial: \( (n-1)! \approx \sqrt{2\pi(n-1)} \left( \frac{n-1}{e} \right)^{n-1} \). Substituting this back gives \( \left( \frac{\pi}{2n} \right)^{n-1} \cdot \sqrt{2\pi(n-1)} \cdot \left( \frac{n-1}{e} \right)^{n-1} \).

Step 5: Simplify the Given Limit

Simplify the limit expression using the fact that \( n \to \infty \): \[ \lim _{n \rightarrow \infty} \left( \frac{(n-1)!}{(2n)^{n-1}} \right)^{1/n} = \lim _{n \rightarrow \infty} \frac{1}{2} \cdot \left( \frac{n-1}{e} \right) \cdot \left( \frac{k!}{(k/e)^k} \right)^{1/n} \].

Step 6: Analyze the Limit Result

The expression now becomes: \( \lim _{n \rightarrow \infty} \left( \frac{n}{2n} \right) = \frac{1}{2} \). Hence, the given expression approaches \( \frac{1}{2} \) as \( n \to \infty \).

Step 7: Correct Answer

In evaluate the limit, the final value found was \( \frac{1}{2} \). Therefore, the correct choice is \((D) \text{None of these}\) since no close answer was presented.

Key Concepts

Stirling's approximation

Stirling's approximation is a powerful tool in mathematics. It is used to approximate the factorial of a large number, which helps simplify complex expressions.

When dealing with factorials, calculating directly can become unmanageable. Stirling's approximation provides a manageable way to estimate factorials, especially as numbers approach infinity.

According to Stirling's approximation, for a large integer \( n \), the factorial \( n! \) can be approximated as:

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

This allows us to transform complicated products into more usable forms for evaluation, such as limits. By using this approximation, solving limit problems becomes much more straightforward.

In the context of calculating limits involving products of sines, applying Stirling's approximation provides a way to handle the factorial terms that emerge. This approximation is crucial for determining growth rates and synthesizing the overall trend as \( n \) grows infinitely large.

Infinite product

An infinite product is a series of multiplicative factors, often represented as a continued multiplication across an infinite sequence. These products can take various forms, and calculating their value, especially as they approach infinity, can be challenging.

In the example given, the problem involves an infinite product of sine functions:

• \( \sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n} \cdot \ldots \cdot \sin \frac{(n-1)\pi}{n} \)

To evaluate the limit of such a product, one common technique is to approximate each term when the argument is very small, using known trigonometric identities.

For small \( x \), \( \sin x \approx x \). This approximation helps in simplifying the expression by transforming it into a sequence more manageable, often reducing the complexity of infinite products to calculations involving polynomials or simple fractions.

Sine function approximation

The sine function, a fundamental trigonometric function, can be approximated under certain conditions to simplify calculations.

This is especially useful when dealing with values that approach a small magnitude, like in limit problems.

For very small angles, the sine function can be approximated using the Taylor series expansion:

• \( \sin x \approx x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots \)
• However, for very small \( x \), the higher-order terms become negligible, leading to \( \sin x \approx x \).

This approximation is particularly useful in evaluating limits involving multiple sine terms, as the simplification from \( \sin x \) to \( x \) reduces the problem to arithmetic that's easier to evaluate.

In our scenario, each term of \( \sin \frac{k\pi}{2n} \) is approximated, allowing complex multiplicative expressions to be expressed as simpler polynomial ones.

Factorial growth rate

Factorial of a number, denoted as \( n! \), represents the product of all positive integers up to \( n \). Factorials grow at a tremendous rate, which is crucial in many areas of mathematical analysis, especially in calculus and combinatorics.

For instance, the rate at which the factorial \( n! \) increases is faster than exponential functions like \( n^n \). This characteristic is a central aspect of solving many limit problems.

In our exercise, the rapid growth of \( (n-1)! \) accompanied with Stirling's approximation plays a key role in solving the limit problem while accounting for the tremendous increase of these terms as \( n \) becomes very large.

Understanding the nature of factorial growth is critical as it impacts predictions and approximations in limits, leading to informed decisions when solving large-scale problems.