Question

Find the limit.

50. \(\mathop {lim}\limits_{x \to 0} \frac{{1 - secx}}{{2x}}\)

Short answer

The value of the limit is 0.

Step-by-step solution

Use the property of limit

A function is continuous at a point “a” if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\)

Evaluate the given limit

Evaluate the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sec x}}{{2x}}{\rm{ }} &= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \frac{1}{{\cos x}}}}{{2x}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\cos x - 1} \right)}}{{\left( {\cos x} \right)\left( {2x} \right)}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{ - \left( {1 - \cos x} \right)}}{{\left( {\cos x} \right)\left( {2x} \right)}}\\ &= - \mathop {\lim }\limits_{x \to 0} \frac{1}{{2\cos x}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{x}} \right)\\ &= - \frac{1}{{2\cos 0}} \times \mathop {\lim }\limits_{x \to 0} \frac{{2.{{\sin }^2}\frac{x}{2}}}{x}\\ &= \frac{{ - 2}}{2} \times \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2}.\frac{{{x^2}}}{4} \times \frac{1}{x}\\ &= - 1 \times 1 \times \mathop {\lim }\limits_{x \to 0} \frac{2}{4}\\ &= 0\end{aligned}\)

Thus, the value of the limit is0.