Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{1 - \sec x}}{{2x}} \cr
& {\text{Use the reciprocal identity }}\sec x = \frac{1}{{\cos x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \frac{1}{{\cos x}}}}{{2x}} \cr
& {\text{Simplifying}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1}}{{2x\cos x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {\frac{{\cos x - 1}}{x}} \right)\left( {\frac{1}{{2\cos x}}} \right)} \right] \cr
& {\text{Use the product law of limits}}{\text{, the limit of a product is the }} \cr
& {\text{product of the limits}}{\text{, then}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos x - 1}}{x}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{2\cos x}}} \right) \cr
& = - \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{x}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{2\cos x}}} \right) \cr
& {\text{Use the special limit }}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{x}} \right) = 0 \cr
& = \left( 0 \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{2\cos x}}} \right) \cr
& = \left( 0 \right)\left( {\frac{1}{{2\cos \left( 0 \right)}}} \right) \cr
& {\text{Simplify}} \cr
& = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{1 - \sec x}}{{2x}} = 0 \cr} $$
