Evaluating the Limit of a Product of Trigonometric Functions

In this article, we will explore the evaluation of the limit (lim_{x to 0} left(frac{3x}{sin 2x} cdot frac{1-cos 6x}{sin^2 x}right)) through a thorough decomposition and application of fundamental limits involving trigonometric functions.

Overview of Trigonometric Limits

First, it's crucial to understand the following fundamental limits for trigonometric functions:

(lim_{x to 0} frac{sin x}{x} 1) (lim_{x to 0} frac{1 - cos x}{x^2} frac{1}{2})

These two fundamental limits are often used to simplify more complex trigonometric limits. We will utilize these as building blocks to solve the given problem.

Step-by-Step Solution

To evaluate the limit, we can separate the given expression into two parts:

(lim_{x to 0} left(frac{3x}{sin 2x} cdot frac{1-cos 6x}{sin^2 x}right) lim_{x to 0} frac{3x}{sin 2x} cdot lim_{x to 0} frac{1-cos 6x}{sin^2 x})

First Limit: (lim_{x to 0} frac{3x}{sin 2x})

By using the limit rule, we can rewrite the first limit as:

(lim_{x to 0} frac{3x}{sin 2x} 3 cdot lim_{x to 0} frac{x}{sin 2x})

Now, let's make a substitution (y 2x). Therefore, as (x to 0), (y to 0) and:

(lim_{x to 0} frac{x}{sin 2x} lim_{y to 0} frac{y/2}{sin y} frac{1}{2} cdot lim_{y to 0} frac{y}{sin y} frac{1}{2} cdot 1 frac{1}{2})

Thus, the first limit simplifies to:

(lim_{x to 0} frac{3x}{sin 2x} 3 cdot frac{1}{2} frac{3}{2})

Second Limit: (lim_{x to 0} frac{1-cos 6x}{sin^2 x})

For the second limit, we can use the trigonometric identity for the cosine function:

(1 - cos 6x 2 sin^2(3x))

Therefore, the second limit becomes:

(lim_{x to 0} frac{1-cos 6x}{sin^2 x} lim_{x to 0} frac{2 sin^2(3x)}{sin^2 x})

Next, let's break it down further:

(lim_{x to 0} frac{2 sin^2(3x)}{sin^2 x} 36 cdot lim_{x to 0} frac{sin(3x)^2}{(3x)^2} cdot lim_{x to 0} frac{(3x)^2}{x^2})

Now, we can use the fundamental limit (lim_{x to 0} frac{sin x}{x} 1) and its derived form:

(lim_{x to 0} frac{sin(3x)}{3x} 1)

Thus:

(lim_{x to 0} frac{(3x)^2}{x^2} 9)

And:

(lim_{x to 0} frac{sin^2(3x)}{(3x)^2} 1)

Multiplying these results together, we get:

(lim_{x to 0} frac{2 sin^2(3x)}{sin^2 x} 36 cdot 1 cdot 9 36 cdot 9 324)

Final Calculation

Combining both parts, we have:

(lim_{x to 0} left(frac{3x}{sin 2x} cdot frac{1-cos 6x}{sin^2 x}right) frac{3}{2} cdot 18 frac{3}{2} cdot 18 frac{39}{2})

Thus, the final answer is:

(boxed{frac{39}{2}})

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