I thought that as $\sec(\pi/2)$ is equal to $1/\cos(\pi/2)$, and given that cos function is positive before $\pi/2$, that it would be positive infinity instead of negative infinity, when $\sec(\pi/2)$ is being approached from the left moving towards the right.
2026-03-28 20:57:28.1774731448
Why is $\sec(\pi/2)$, when approaching from the left equal to negative infinity and not positive infinity?
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As you said "... cos function is positive before pi/2 ..." and "before pi/2" means you are less than pi/2, you are on the "right" side of pi/2 considering unit circle! So, if you approach pi/2 "from right" you get infinity.
Simiarly if you approach pi/2 "from left", you are on the left side of pi/2 considering unit circle and since cos function is negative there, you get minus infinity.
However, on the real line you are right that,
"$\pi/2$ being approached from the left moving towards the right" is denoted by $\displaystyle \lim_{x\to (\pi/2)^-}$ which means $x<\pi/2$ and $x \to \pi/2$.
"$\pi/2$ being approached from the right moving towards the left" is denoted by $\displaystyle \lim_{x\to (\pi/2)^+}$ which means $x>\pi/2$ and $x \to \pi/2$.