Calculus 8th Edition

$-\infty$
$x\rightarrow(\pi/2)^{+}$ means that x approaches $\displaystyle \frac{\pi}{2}$ from the right (the greater side), it is positive, so $\displaystyle \frac{1}{x}$ is positive. Secant and cosine are reciprocal. The cosine (and secant) of x, belonging (slightly) to quadrant II, are negative. Since $\cos x$ approaches 0 from the negative side, $\sec x\rightarrow-\infty$ So, as $x\rightarrow(\pi/2)^{+}$ the product ($\displaystyle \frac{1}{x}\sec x$) $\rightarrow-\infty$.