Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 21

$2$

Let $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{x^2}{\ln (\sec x)}$ But, $f(0)=\dfrac{0^2}{\ln (\sec 0)}=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{x \to 0}\dfrac{2}{\sec^2 x}=\dfrac{2}{\sec^20}=2$