Evaluating The Limit Of (1 - Cos Θ) / Sin 2θ As Θ Approaches 0

This article delves into the evaluation of a specific limit problem encountered in calculus, focusing on the expression ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta}}$. This limit problem is a classic example often used to illustrate the application of trigonometric identities and limit properties. Understanding how to solve this type of problem is crucial for mastering calculus and related mathematical concepts. We will explore various methods to approach this limit, including the use of trigonometric identities, L'Hôpital's Rule, and small-angle approximations. Each method provides a unique perspective on the problem, enhancing our understanding of limits and trigonometric functions.

Understanding the Limit Problem

At its core, the limit ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta}}$ asks us to determine the behavior of the function ${f(\theta) = \frac{1-\cos \theta}{\sin 2 \theta}}$ as ${\theta}$ gets arbitrarily close to 0. Directly substituting ${\theta = 0}$ into the function results in an indeterminate form of ${\frac{0}{0}}$, as both the numerator (1 - cos 0 = 1 - 1 = 0) and the denominator (sin(2 * 0) = sin 0 = 0) become zero. This indeterminate form signals that we cannot simply plug in the value and need to employ other techniques to find the limit. Indeterminate forms are common in calculus, and various strategies have been developed to handle them. Recognizing the indeterminate form is the first step in choosing the appropriate method for evaluating the limit. The expression highlights the interplay between trigonometric functions and the concept of limits, making it an excellent exercise for reinforcing fundamental calculus principles. To accurately determine the limit, one must manipulate the expression to eliminate the indeterminate form. This usually involves applying trigonometric identities, algebraic manipulations, or more advanced techniques like L'Hôpital's Rule.

Method 1: Using Trigonometric Identities

One effective approach to evaluate this limit involves leveraging trigonometric identities. The key identity to consider is the double-angle identity for sine, which states that ${\sin 2\theta = 2 \sin \theta \cos \theta}$. Additionally, we can use the identity ${1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})}$. By applying these identities, we can rewrite the original expression in a more manageable form. The double-angle identity allows us to express ${\sin 2\theta}$ in terms of ${\sin \theta}$ and ${\cos \theta}$, while the identity for ${1 - \cos \theta}$ introduces a squared sine term, which can be helpful in simplifying the fraction. Substituting these identities into the limit expression, we get:

$$\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} = \lim _{\theta \rightarrow 0} \frac{2 \sin^2(\frac{\theta}{2})}{2 \sin \theta \cos \theta}$$

Now, we can further manipulate the expression to introduce terms that are familiar in the context of limits. Specifically, we aim to utilize the fundamental trigonometric limit ${\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$. To achieve this, we can rewrite ${\sin \theta}$ as ${2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})}$. This substitution is derived from the double-angle identity applied in reverse. By expressing ${\sin \theta}$ in terms of half-angle sine and cosine, we create an opportunity to utilize the ${\frac{\sin x}{x}}$ limit. Substituting this back into the expression, we obtain:

$$\lim _{\theta \rightarrow 0} \frac{2 \sin^2(\frac{\theta}{2})}{2 (2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})) \cos \theta} = \lim _{\theta \rightarrow 0} \frac{\sin^2(\frac{\theta}{2})}{2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) \cos \theta}$$

We can now cancel out a ${\sin(\frac{\theta}{2})}$ term from the numerator and denominator, provided that ${\sin(\frac{\theta}{2}) \neq 0}$, which is valid as ${\theta}$ approaches 0 but is not exactly 0. This simplification leads to:

$$\lim _{\theta \rightarrow 0} \frac{\sin(\frac{\theta}{2})}{2 \cos(\frac{\theta}{2}) \cos \theta}$$

Now, we can evaluate the limit by direct substitution. As ${\theta}$ approaches 0, ${\sin(\frac{\theta}{2})}$ approaches 0, ${\cos(\frac{\theta}{2})}$ approaches 1, and ${\cos \theta}$ approaches 1. Therefore, the limit becomes:

$$\frac{0}{2 * 1 * 1} = 0$$

Thus, using trigonometric identities, we find that ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} = 0}$.

Method 2: Applying L'Hôpital's Rule

Another powerful technique for evaluating limits of indeterminate forms is L'Hôpital's Rule. This rule states that if the limit of ${\frac{f(\theta)}{g(\theta)}}$ as ${\theta}$ approaches a value results in an indeterminate form such as ${\frac{0}{0}}$ or ${\frac{\infty}{\infty}}$, then the limit can be found by taking the derivatives of the numerator and the denominator separately and then evaluating the limit of the new quotient, provided that the limit exists. Mathematically, if ${\lim_{\theta \rightarrow c} f(\theta) = 0}$ and ${\lim_{\theta \rightarrow c} g(\theta) = 0}$ (or both limits are infinite), and if ${\lim_{\theta \rightarrow c} \frac{f'(\theta)}{g'(\theta)}}$ exists, then:

$$\lim _{\theta \rightarrow c} \frac{f(\theta)}{g(\theta)} = \lim _{\theta \rightarrow c} \frac{f'(\theta)}{g'(\theta)}$$

In our case, we have ${f(\theta) = 1 - \cos \theta}$ and ${g(\theta) = \sin 2\theta}$. As we saw earlier, substituting ${\theta = 0}$ directly into the function yields the indeterminate form ${\frac{0}{0}}$, making L'Hôpital's Rule applicable. To apply the rule, we first need to find the derivatives of ${f(\theta)}$ and ${g(\theta)}$ with respect to ${\theta}$. The derivative of ${f(\theta) = 1 - \cos \theta}$ is:

$$f'(\theta) = \frac{d}{d\theta}(1 - \cos \theta) = 0 - (-\sin \theta) = \sin \theta$$

The derivative of ${g(\theta) = \sin 2\theta}$ is:

$$g'(\theta) = \frac{d}{d\theta}(\sin 2\theta) = 2 \cos 2\theta$$

Now, we apply L'Hôpital's Rule by taking the limit of the quotient of these derivatives:

$$\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} = \lim _{\theta \rightarrow 0} \frac{\sin \theta}{2 \cos 2\theta}$$

Next, we try to evaluate this new limit by direct substitution. As ${\theta}$ approaches 0, ${\sin \theta}$ approaches 0, and ${\cos 2\theta}$ approaches ${\cos(0) = 1}$. Therefore, the limit becomes:

$$\frac{0}{2 * 1} = 0$$

Thus, by applying L'Hôpital's Rule, we again find that ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} = 0}$. This method offers a straightforward approach for solving limits of indeterminate forms, especially when dealing with trigonometric functions.

Method 3: Utilizing Small-Angle Approximations

When dealing with limits as ${\theta}$ approaches 0, we can employ small-angle approximations to simplify trigonometric functions. For small values of ${\theta}$ (in radians), the following approximations hold:

• ${\sin \theta \approx \theta}$
• ${\cos \theta \approx 1 - \frac{\theta^2}{2}}$

These approximations stem from the Taylor series expansions of sine and cosine functions around ${\theta = 0}$. The smaller the value of ${\theta}$, the more accurate these approximations become. In the context of our limit problem, we can use these approximations to simplify the expression ${\frac{1-\cos \theta}{\sin 2 \theta}}$. Substituting the small-angle approximations, we get:

$$\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} \approx \lim _{\theta \rightarrow 0} \frac{1 - (1 - \frac{\theta^2}{2})}{2\theta}$$

Here, we have approximated ${\sin 2\theta}$ as ${2\theta}$, using the small-angle approximation for sine. Now, we simplify the numerator:

$$\lim _{\theta \rightarrow 0} \frac{1 - (1 - \frac{\theta^2}{2})}{2\theta} = \lim _{\theta \rightarrow 0} \frac{\frac{\theta^2}{2}}{2\theta}$$

Further simplification involves dividing ${\frac{\theta^2}{2}}$ by ${2\theta}$:

$$\lim _{\theta \rightarrow 0} \frac{\theta^2}{4\theta} = \lim _{\theta \rightarrow 0} \frac{\theta}{4}$$

Now, we can directly substitute ${\theta = 0}$ into the simplified expression:

$$\frac{0}{4} = 0$$

Therefore, using small-angle approximations, we again arrive at the conclusion that ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} = 0}$. This method provides a quick and intuitive way to evaluate the limit, especially when dealing with trigonometric functions as the variable approaches zero. However, it is important to remember that these approximations are valid only for small values of ${\theta}$.

Conclusion

In summary, we have explored three distinct methods to evaluate the limit ${\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta}}$: trigonometric identities, L'Hôpital's Rule, and small-angle approximations. Each method provides a unique approach and reinforces fundamental calculus principles. By applying trigonometric identities, we manipulated the expression to utilize the fundamental trigonometric limit ${\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$, leading to the solution. L'Hôpital's Rule offered a direct way to handle the indeterminate form by differentiating the numerator and denominator. Small-angle approximations provided a simplified approach by approximating trigonometric functions for small values of ${\theta}$. All three methods consistently demonstrated that the limit evaluates to 0. This exercise highlights the importance of understanding various techniques for limit evaluation and the power of trigonometric identities, L'Hôpital's Rule, and approximations in simplifying complex expressions. Mastering these methods is crucial for success in calculus and related fields.