the function, f(x) = sec(x), is plotted as shown. for which values of x is the limit positive infinity when…

Explanation:

Step1: Recall secant - cosine relationship

Recall that $\sec(x)=\frac{1}{\cos(x)}$. We want to find when $\lim_{x\rightarrow a^{-}}\sec(x)=+\infty$, which is equivalent to $\lim_{x\rightarrow a^{-}}\cos(x)=0^{+}$ (since $\frac{1}{0^{+}}=+\infty$).

Step2: Analyze the cosine - function behavior

The cosine function $y = \cos(x)$ has a value of 0 at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$. When $x\rightarrow-\frac{3\pi}{2}^{-}$, $\cos(x)\rightarrow0^{+}$ because as $x$ approaches $-\frac{3\pi}{2}$ from the left, the values of $\cos(x)$ approach 0 through positive values. When $x\rightarrow\frac{\pi}{2}^{-}$, $\cos(x)\rightarrow0^{+}$ because as $x$ approaches $\frac{\pi}{2}$ from the left, the values of $\cos(x)$ approach 0 through positive values.

Answer:

$\frac{\pi}{2},-\frac{3\pi}{2}$