We are tasked with finding the limit:

$\lim_{\theta\to 0}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}.$

Step 1: Use trigonometric identities and approximations

As $\theta$ approaches 0, we can use the following approximations for small angles:

• $\sin(\theta) \approx \theta$
• $1 - \cos(\theta) \approx \dfrac{\theta^2}{2}$

Step 2: Apply the approximations

Using the approximation $1 - \cos(\theta) \approx \dfrac{\theta^2}{2}$ and $\sin^2(\theta) \approx \theta^2$, we substitute into the given limit expression:

$\dfrac{1-\cos(\theta)}{2\sin^2(\theta)} \approx \dfrac{\dfrac{\theta^2}{2}}{2\theta^2}.$

Step 3: Simplify the expression

Simplifying the above expression:

$\dfrac{\dfrac{\theta^2}{2}}{2\theta^2} = \dfrac{1}{4}.$

Conclusion:

Thus, the limit is:

$\lim_{\theta\to 0}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)} = \dfrac{1}{4}.$

The correct answer is Choice C: $\dfrac{1}{4}$.

Would you like more details or explanations? Here are 5 related questions:

1. What is L'Hôpital's Rule and when is it useful?
2. How can we apply Taylor series to approximate trigonometric functions?
3. What other limits are useful to memorize for small angle approximations?
4. What is the formal definition of a limit in calculus?
5. How does $\sin(\theta) \approx \theta$ for small $\theta$ relate to the unit circle?

Tip: When dealing with limits of trigonometric functions, always check if small angle approximations can simplify the problem!