Do I need to draw triangles as well?

$\lim_{\theta \to 0} \frac{1 - \cos \theta}{2(1 - \cos^2 \theta)} = \\\\ \lim_{\theta \to 0} \frac{1 - \cos \theta}{2(1 + \cos \theta)(1 - \cos \theta)} = \\\\ \frac{1}{2(1 + 1)} = \\\\ \boxed{\frac{1}{4}}$

The limit has a $\frac{0}{0}$ indeterminate form. Apply L'Hospital Rule to get, $\frac{sin \theta}{2sin 2\theta}$ , which is again in $\frac{0}{0}$ indeterminate form. Apply L'Hospital rule once again to obtain $\frac{cos \theta}{4cos 2\theta}$ . Since cos is continuous functions, we can plug in the limit value of $\theta = 0$ into the expression.

$we\quad can\quad change\quad the\quad numerator\quad \left( 1-\cos { \theta } \right) \quad into\quad 2\sin ^{ 2 }{ \frac { \theta }{ 2 } }$

$\Rightarrow \frac { 2\sin ^{ 2 }{ \frac { \theta }{ 2 } } }{ 2\sin ^{ 2 }{ \theta } }$

$Now\quad if\quad we\quad multiply\quad and\quad divide\quad the\quad numerator\quad by\quad \frac { { \theta }^{ 2 } }{ 4 } \quad and\quad denominator\quad by\quad { \theta }^{ 2\quad }we\quad get$

$\Rightarrow \frac { 2\sin ^{ 2 }{ \frac { \theta }{ 2 } \left( \frac { { \theta }^{ 2 } }{ 4 } \right) } \times \left( { \theta }^{ 2 } \right) }{ \left( \frac { { \theta }^{ 2 } }{ 4 } \right) \times 2\sin ^{ 2 }{ \theta .\left( { \theta }^{ 2 } \right) } }$

$Now\quad applying\quad the\quad property\\ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } =1 } \quad$

$We\quad get\\ \lim _{ \theta \rightarrow 0 }{ \frac { { \theta }^{ 2 } }{ 4{ \theta }^{ 2 } } }$

$\boxed { =\frac { 1 }{ 4 } }$

Multiply top and bottom by 1+cos(θ) to end up with limθ→01−cos2(θ)/2sin2(θ)(1+cos(θ))=limθ→01/2(1+cos(θ))=1/4 where the identity 1−cos2(θ)=sin2(θ) was used.

Multiply top and bottom by $1 + \cos(\theta)$ to end up with

$\lim_{\theta \rightarrow 0} \dfrac{1 - \cos^{2}(\theta)}{2\sin^{2}(\theta)(1 + \cos(\theta))} = \lim_{\theta \rightarrow 0} \dfrac{1}{2(1 + \cos(\theta))} = \boxed{\dfrac{1}{4}},$

where the identity $1 - \cos^{2}(\theta) = \sin^{2}(\theta)$ was used.