Proving the Limit of n sin(π/n) π Using Advanced Techniques

In this article, we will delve into the fascinating world of mathematical proofs, specifically focusing on the limit limn→∞ n sin(π/n) π. We will explore various methods to prove this intriguing result, emphasizing the use of limits and the small-angle approximation for the sine function. By the end of this article, you will gain a deeper understanding of how to approach and solve such problems.

Introduction

The limit in question is a classic example of a problem that requires a sophisticated approach to solve. It combines elements of calculus and trigonometry, making it a perfect topic for advanced learners and professionals seeking to sharpen their analytical skills.

Step 1: Substitution Method

Let's begin by substituting x π/n. As n → ∞, x → 0. This substitution allows us to rewrite the limit in terms of x as follows:

limn→∞ n sin(π/n) limx→0 (π/x) sin(x)

Step 2: Simplification and Small-Angle Approximation

Now, we simplify the expression (π/x) sin(x). A fundamental limit from calculus states that:

limx→0 sin(x)/x 1

Using this, we can approximate sin(x) ≈ x for small values of x. Therefore:

sin(x) ≈ x - o(x)

where o(x) represents terms that go to zero faster than x.

Substituting back, we get:

(π/x) sin(x) ≈ (π/x) [x - o(x)] π - (π/x) o(x)

As x → 0, (π/x) o(x) → 0, leading to:

limx→0 (π - (π/x) o(x)) π

Alternative Approach

A second approach involves a change of variables. Let's set a π/n. The limit then becomes:

limn→∞ n sin(π/n) lima→0^ (π/a) sin(a) π lima→0^ (sin(a)/a) π

Since it is well-known that:

lima→0 (sin(a)/a) 1

We can directly substitute:

limn→∞ n sin(π/n) π * 1 π

Using the Limit Transformation Technique

Another method involves transforming the limit into a more manageable form. Consider:

limn→∞ n sin(π/n) limx→0 (π/x) sin(x) π limx→0 (sin(x)/x) π * 1 π

Here, we have used the transformation n π/x and the well-known limit:

limx→0 (sin(x)/x) 1

Laurent Series Expansion

For a more rigorous and detailed proof, we can use the Laurent series expansion. The Taylor series expansion for sine around zero is:

sin(x) x - (x^3)/6 o(x^3)

Substituting this into the limit:

limn→∞ n sin(π/n) limn→∞ [n(π/n - (π/n)^3/6 o((π/n)^3))] limn→∞ [π - (π^3)/(6n^2) o(1/n^2)]

As n → ∞, the higher-order terms approach zero:

limn→∞ [π - (π^3)/(6n^2) o(1/n^2)] π

Thus, we have:

limn→∞ n sin(π/n) π

Conclusion

In conclusion, we have explored multiple methods to prove the limit limn→∞ n sin(π/n) π. The techniques range from substitution and small-angle approximations to Laurent series expansions. Each method provides a unique perspective on solving this limit problem, offering valuable insights into the interplay between trigonometric functions and limits.