Why is $\sec(\pi/2)$, when approaching from the left equal to negative infinity and not positive infinity?

$\begingroup$

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.

$\endgroup$ 4

1 Answer

$\begingroup$

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 $x \to \pi/2$.

$\endgroup$ 5

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy