Answer is: option3

Solution:

Use L’Hospital’s Rule; then

\[ \lim_{x \to 0} \frac{\sec x - \cos x}{x^2} = \lim_{x \to 0} \frac{\sec x \tan x + \sin x}{2x} \]

\[ = \lim_{x \to 0} \frac{\sec^3 x + \sec x \tan^2 x + \cos x}{2} = \frac{1 + 1 \cdot 0 + 1}{2} = 1 \]

This can be done without using L’Hospital’s Rule as follows:

\[ \lim_{x \to 0} \frac{\sec x - \cos x}{x^2} = \lim_{x \to 0} \left( \frac{1}{\cos x} - \cos x \right) \frac{1}{x^2} \]

\[ = \lim_{x \to 0} \frac{1 - \cos^2 x}{x^2 \cos x} \]

\[ = \lim_{x \to 0} \frac{\sin^2 x}{x^2 \cos x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 \cdot 1 = 1 \]

Final Answer: 1