To find the limit [tex]\(\lim _{\theta \rightarrow 0} \frac{1-\cos (\theta)}{2 \sin ^2(\theta)}\)[/tex], we observe that substituting [tex]\(\theta = 0\)[/tex] directly into the expression results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex]. Therefore, we need to apply L'Hôpital's Rule, which helps solve limits of indeterminate forms.
L'Hôpital's Rule states that if [tex]\(\lim_{\theta \to a} \frac{f(\theta)}{g(\theta)}\)[/tex] results in an indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[
\lim_{\theta \to a} \frac{f(\theta)}{g(\theta)} = \lim_{\theta \to a} \frac{f'(\theta)}{g'(\theta)}
\][/tex]
provided this limit exists.
First, let's identify the functions in our expression:
[tex]\[
f(\theta) = 1 - \cos(\theta)
\][/tex]
[tex]\[
g(\theta) = 2 \sin^2(\theta)
\][/tex]
Next, we find the derivatives of [tex]\(f(\theta)\)[/tex] and [tex]\(g(\theta)\)[/tex].
For the numerator [tex]\(f(\theta) = 1 - \cos(\theta)\)[/tex]:
[tex]\[
f'(\theta) = \frac{d}{d\theta} (1 - \cos(\theta)) = \sin(\theta)
\][/tex]
For the denominator [tex]\(g(\theta) = 2 \sin^2(\theta)\)[/tex]:
[tex]\[
g'(\theta) = \frac{d}{d\theta} (2 \sin^2(\theta)) = 2 \cdot 2 \sin(\theta) \cos(\theta) = 4 \sin(\theta) \cos(\theta)
\][/tex]
Now we use L'Hôpital's Rule:
[tex]\[
\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{2 \sin^2(\theta)} = \lim_{\theta \to 0} \frac{\sin(\theta)}{4 \sin(\theta) \cos(\theta)}
\][/tex]
We simplify the expression:
[tex]\[
\lim_{\theta \to 0} \frac{\sin(\theta)}{4 \sin(\theta) \cos(\theta)} = \lim_{\theta \to 0} \frac{1}{4 \cos(\theta)}
\][/tex]
As [tex]\(\theta\)[/tex] approaches 0, [tex]\(\cos(\theta)\)[/tex] approaches 1. So:
[tex]\[
\lim_{\theta \to 0} \frac{1}{4 \cos(\theta)} = \frac{1}{4 \cdot 1} = \frac{1}{4}
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{\frac{1}{4}}
\][/tex]
