Proving the Limit of a Trigonometric Product

Understanding and proving the limit of complex trigonometric products can often be a challenging task, but it is also a fundamental skill in calculus and analysis. In this article, we will explore how to prove the following limit:

Problem Statement

Consider the limit of the product of tangent functions as follows:

(lim_{n to infty} left(prod_{k1}^{n-1} tanleft(frac{kpi}{2n}right)right)^{frac{1}{n}} 1)

Step-by-Step Proof

Let's break down the proof using natural logarithms and some trigonometric identities.

Step 1: Applying Natural Logarithm

First, we take the natural logarithm of the given expression:

(lnleft(lim_{n to infty} left(prod_{k1}^{n-1} tanleft(frac{kpi}{2n}right)right)^{frac{1}{n}}right) ln L)

Step 2: Rearranging the Limit

Next, we rearrange the limit and the natural logarithm:

(lim_{n to infty} frac{1}{n} ln left(prod_{k1}^{n-1} tanleft(frac{kpi}{2n}right)right) ln L)

Step 3: Converting the Product into a Sum

Using the property of logarithms, we can convert the product into a sum:

(lim_{n to infty} frac{1}{n} sum_{k1}^{n-1} ln left(tanleft(frac{kpi}{2n}right)right) ln L)

Step 4: Utilizing Trigonometric Identities

Recall the identity for tangent and cotangent:

(tan x frac{sin x}{cos x})

Using this identity, we can rewrite the expression inside the limit:

(lim_{n to infty} frac{1}{n} sum_{k1}^{n-1} ln left(tanleft(frac{kpi}{2n}right)right) lim_{n to infty} left[frac{1}{n} sum_{k1}^{n-1} ln left(frac{sin left(frac{kpi}{2n}right)}{cos left(frac{kpi}{2n}right)}right)right])

Which simplifies to:

(lim_{n to infty} left[frac{1}{n} sum_{k1}^{n-1} ln left(sinleft(frac{kpi}{2n}right)right) - frac{1}{n} sum_{k1}^{n-1} ln left(cosleft(frac{kpi}{2n}right)right)right])

Using the identity for the cosine function, we get:

(cos left(frac{kpi}{2n}right) sin left(frac{pi}{2} - frac{kpi}{2n}right) sin left(frac{n-kpi}{2n}right))

So the expression becomes:

(lim_{n to infty} left[frac{1}{n} sum_{k1}^{n-1} ln left(sin left(frac{kpi}{2n}right)right) - frac{1}{n} sum_{k1}^{n-1} ln left(sin left(frac{n-kpi}{2n}right)right)right])

Notice that the second sum is the first sum in reverse, but since the logarithm is odd around 0, the entire difference is zero:

(lim_{n to infty} left[frac{1}{n} sum_{k1}^{n-1} ln left(sin left(frac{kpi}{2n}right)right) - frac{1}{n} sum_{k1}^{n-1} ln left(sin left(frac{n-kpi}{2n}right)right)right] 0)

Therefore, we have:

(ln L 0 Rightarrow L 1)

Alternative Proof

Alternatively, we can use a simpler approach. Consider the product:

(A_n prod_{k1}^{n-1} tan left(frac{kpi}{2n}right))

By replacing k with n-k, we get:

(A_n prod_{k1}^{n-1} tan left(frac{n-kpi}{2n}right) prod_{k1}^{n-1} cot left(frac{kpi}{2n}right))

This implies:

(A_n^2 prod_{k1}^{n-1} tan left(frac{kpi}{2n}right) cot left(frac{kpi}{2n}right) 1)

Thus, we conclude that:

(A_n 1) for every n

Therefore, the limit of the original expression as n approaches infinity is:

(lim_{n to infty} A_n^{frac{1}{n}} 1)

Conclusion

The proof shows that the limit of the given trigonometric product is indeed 1. This result highlights the power of using natural logarithms and trigonometric identities in solving complex limit problems.

For further reading on similar topics, we recommend exploring the properties of trigonometric functions and the application of limits in calculus. Understanding these concepts will enhance your mathematical toolkit and problem-solving abilities.