Answer
(a)$$D_f= \mathbb{R}- \left ( \{ 0 \} \cup \{ x \mid x=(2n+1) \frac{\pi }{2}, \, n \in \mathbb{Z} \} \right )$$
(b) Please see below.
(c)$$\lim_{x \to 0 } f(x)=\frac{1}{2}$$
(d) Please see below.
Work Step by Step
(a) In fact, there exist two ratios in the function $f(x)= \frac{ \sec x -1}{x^2}=\frac{\frac{1}{\cos x}-1}{x^2}$. So, by excluding the points vanishing the denominators we can find the domain of $f$ as follows.$$x^2=0 \quad \Rightarrow \quad x=0 \\ \cos x=0 \quad \Rightarrow \quad x= (2n+1)\frac{\pi }{2}, \quad n\in \mathbb{Z}$$Thus, the domain of $f$ is$$D_f= \mathbb{R}- \left ( \{ 0 \} \cup \{ x \mid x=(2n+1) \frac{\pi }{2}, \, n \in \mathbb{Z} \} \right ).$$
(b) No, it is not obvious from the graph since the graph seems to be a graph of a periodic function and so consists of infinitely many pieces. However, the graph confirms our result.
(c) According to the graph, as $x$ approaches $0$, the function $f(x)$ approaches $\frac{1}{2}$. So, we can conclude that$$\lim_{x \to 0}f(x)=0.$$
(d) Now, we find the limit analytically. Since by direct substitution we get the indeterminate form $\frac{0}{0}$, we should use the trigonometric relation $\cos x= 1- 2 \sin ^2 \frac{x}{2}$ as folllows.$$\lim_{x \to 0}\frac {\sec x -1}{x^2}=\lim_{x \to 0} \frac{\frac{1}{\cos x}- 1}{x^2}= \lim_{x \to 0}\frac{\frac{1-\cos x}{\cos x}}{x^2}=\lim_{x \to 0}\left ( \frac{1}{\cos x} \right ) \left (\frac{1- \cos x}{x^2} \right )=\lim_{x \to 0}\left ( \frac{1}{\cos x } \right )\left (\frac{2\sin ^2 \frac{x}{2}}{x^2} \right )=\lim_{x \to 0} \left ( \frac{1}{2 \cos x} \right ) \left ( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right )^2= \lim_{u \to 0} \left ( \frac{1}{2 \cos 2u} \right ) \left ( \frac{\sin u}{u} \right )^2= ( \frac{1}{2}) (1)^2=\frac{1}{2}$$(In finding the limit we have used Thereom 1.9 (a), $\lim_{x \to 0} \frac{\sin x}{x}=1$, also $u=\frac{x}{2}$).
