Evaluating Limits A Step-by-Step Solution For Lim (x->0) (x - X Cos X) / Sin^2(3x)

In the realm of calculus, limits serve as a cornerstone for understanding the behavior of functions as their input approaches a specific value. The limit, denoted as lim (x->c) f(x), essentially asks: What value does f(x) get arbitrarily close to as x gets arbitrarily close to c? Evaluating limits often involves a blend of algebraic manipulation, trigonometric identities, and the application of powerful tools like L'Hôpital's Rule. This article delves into the intricacies of evaluating the limit lim (x->0) (x - x cos x) / sin^2(3x), providing a comprehensive exploration of the techniques involved and the underlying concepts.

Decoding the Limit Expression

At first glance, the expression (x - x cos x) / sin^2(3x) might seem daunting. However, by strategically applying trigonometric identities and algebraic manipulations, we can unveil its hidden structure and pave the way for evaluation. Our primary goal is to transform the expression into a form where we can directly apply limit laws or recognize standard limit results. Let's begin by factoring out x from the numerator, which gives us x(1 - cos x) / sin^2(3x). This seemingly simple step is crucial, as it isolates a key component, (1 - cos x), which we can further manipulate using a well-known trigonometric identity.

To tackle the (1 - cos x) term, we employ the identity 1 - cos x = 2sin^2(x/2). This identity allows us to rewrite the numerator as x * 2sin^2(x/2). Now our expression transforms to 2xsin^2(x/2) / sin^2(3x). This form is more amenable to analysis, as it highlights the sine functions, which have predictable behavior as x approaches 0. The next step involves strategically manipulating the denominator to align it with the sine terms in the numerator. Recall the standard limit lim (x->0) sin(x)/x = 1. This limit will be our guiding principle as we reshape the expression.

To leverage this standard limit, we aim to introduce terms that resemble sin(x)/x. In the denominator, we have sin^2(3x). We can rewrite this as (sin(3x))^2. To introduce the desired form, we multiply and divide by (3x)^2, which gives us (sin(3x))^2 / (3x)^2 * (3x)^2. Now, we have the term (sin(3x) / 3x)^2, which approaches 1 as x approaches 0. The remaining factor, (3x)^2, simplifies to 9x^2. Substituting this back into our original expression, we get 2xsin^2(x/2) / (9x^2 * (sin(3x) / 3x)^2). This transformation brings us closer to a form where we can directly apply the limit laws.

Applying Limit Laws and Standard Results

Now that we have transformed the expression into 2xsin^2(x/2) / (9x^2 * (sin(3x) / 3x)^2), we can apply limit laws and standard results to evaluate the limit as x approaches 0. The expression can be further simplified by canceling out a factor of x from the numerator and denominator, yielding 2sin^2(x/2) / (9x * (sin(3x) / 3x)^2). At this point, it's crucial to remember the limit laws, which allow us to break down the limit of a quotient into the quotient of limits, provided the limits in the denominator do not approach zero. However, in our case, the denominator still contains a term of x, which approaches 0 as x approaches 0. Therefore, we need to proceed with caution and further manipulation.

To address the remaining x term in the denominator, we once again turn to the standard limit lim (x->0) sin(x)/x = 1. We have sin^2(x/2) in the numerator. To create the desired form, we multiply and divide by (x/2)^2, which gives us sin^2(x/2) / (x/2)^2 * (x/2)^2. Now we have the term (sin(x/2) / (x/2))^2, which approaches 1 as x approaches 0. The remaining factor, (x/2)^2, simplifies to x^2/4. Substituting this back into our expression, we get 2 * (sin^2(x/2) / (x/2)^2) * (x^2/4) / (9x * (sin(3x) / 3x)^2). This manipulation allows us to isolate the standard limit and simplify the expression further.

We can now cancel out a factor of x from the numerator and denominator, and rearrange the terms to get (2 * (sin^2(x/2) / (x/2)^2) * (x/4)) / (9 * (sin(3x) / 3x)^2). As x approaches 0, the term (sin^2(x/2) / (x/2)^2) approaches 1, and the term (sin(3x) / 3x)^2 also approaches 1. The remaining term in the numerator, x/4, approaches 0 as x approaches 0. Therefore, the entire expression approaches (2 * 1 * 0) / (9 * 1), which simplifies to 0. Thus, we conclude that the limit of the given expression as x approaches 0 is 0.

L'Hôpital's Rule: An Alternative Approach

While we successfully evaluated the limit using trigonometric identities and standard results, it's worth exploring an alternative approach using L'Hôpital's Rule. L'Hôpital's Rule provides a powerful technique for evaluating limits of indeterminate forms, such as 0/0 or ∞/∞. Our original limit, lim (x->0) (x - x cos x) / sin^2(3x), falls into the 0/0 indeterminate form, as both the numerator and denominator approach 0 as x approaches 0. L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches c is of the form 0/0 or ∞/∞, and if the limit of f'(x)/g'(x) exists, then lim (x->c) f(x)/g(x) = lim (x->c) f'(x)/g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively.

To apply L'Hôpital's Rule, we need to find the derivatives of the numerator and the denominator. Let f(x) = x - x cos x and g(x) = sin^2(3x). The derivative of f(x), f'(x), is found using the product rule: f'(x) = 1 - (cos x - x sin x) = 1 - cos x + x sin x. The derivative of g(x), g'(x), is found using the chain rule: g'(x) = 2sin(3x) * cos(3x) * 3 = 6sin(3x)cos(3x). Now we have the limit of f'(x)/g'(x) as x approaches 0, which is lim (x->0) (1 - cos x + x sin x) / (6sin(3x)cos(3x)). This limit is still in an indeterminate form of 0/0, as both the numerator and denominator approach 0 as x approaches 0. Therefore, we can apply L'Hôpital's Rule again.

We need to find the second derivatives of f(x) and g(x). The derivative of f'(x), f''(x), is found as follows: f''(x) = sin x + (sin x + x cos x) = 2sin x + x cos x. The derivative of g'(x), g''(x), is found using the product rule and chain rule: g''(x) = 6(cos(3x) * 3 * cos(3x) + sin(3x) * (-sin(3x) * 3)) = 18(cos^2(3x) - sin^2(3x)) = 18cos(6x). Now we have the limit of f''(x)/g''(x) as x approaches 0, which is lim (x->0) (2sin x + x cos x) / (18cos(6x)). As x approaches 0, the numerator approaches 2sin(0) + 0 * cos(0) = 0, and the denominator approaches 18cos(6*0) = 18cos(0) = 18. Therefore, the limit is 0/18, which simplifies to 0. This result confirms our earlier finding using trigonometric identities and standard limits. L'Hôpital's Rule provides a powerful alternative method for evaluating limits of indeterminate forms, and its application in this case reinforces our conclusion that the limit of the given expression as x approaches 0 is indeed 0.

Conclusion: A Multifaceted Approach to Limit Evaluation

In this exploration, we have meticulously evaluated the limit lim (x->0) (x - x cos x) / sin^2(3x) using two distinct approaches: trigonometric identities and standard limits, and L'Hôpital's Rule. Both methods converged to the same result, solidifying our understanding that the limit is 0. The first approach involved a series of algebraic and trigonometric manipulations, leveraging the identity 1 - cos x = 2sin^2(x/2) and the standard limit lim (x->0) sin(x)/x = 1. This method highlighted the power of strategic transformations and the importance of recognizing standard limit results.

The second approach employed L'Hôpital's Rule, a powerful tool for evaluating limits of indeterminate forms. By repeatedly applying the rule, we were able to bypass the complexities of trigonometric manipulations and arrive at the same conclusion. This demonstrates the versatility of calculus techniques and the value of having multiple tools at one's disposal. The evaluation of limits is a fundamental concept in calculus, and mastering various techniques for evaluating them is crucial for success in more advanced topics. By understanding the interplay between algebraic manipulation, trigonometric identities, standard limits, and L'Hôpital's Rule, we can confidently tackle a wide range of limit problems.

This detailed analysis not only provides a solution to the specific limit problem but also serves as a valuable learning experience, showcasing the diverse strategies and techniques employed in calculus. The ability to approach a problem from multiple angles and adapt one's approach based on the specific characteristics of the problem is a hallmark of mathematical proficiency. As we continue our journey through calculus, the insights gained from this exploration will undoubtedly serve us well in tackling more complex and challenging problems.

Find the limit of (x - x cos x) / sin^2(3x) as x approaches 0.

Evaluating Limits A Step-by-Step Solution for lim (x->0) (x - x cos x) / sin^2(3x)