question on arcsec infinity

**FailCalculus** · July 31st 2010, 07:50 AM

can someone explain to me how arcsec infinity is equal to pi/2? thanks

**mathaddict** · July 31st 2010, 07:58 AM

Originally Posted by FailCalculus

can someone explain to me how arcsec infinity is equal to pi/2? thanks

$\sec \theta=\frac{1}{\cos \theta}$

Set the denominator to be 0

**HallsofIvy** · July 31st 2010, 11:35 AM

Originally Posted by FailCalculus

can someone explain to me how arcsec infinity is equal to pi/2? thanks

Strictly speaking, " $arcsec(\infty)$ " isn't $\pi/2$ or any other number. " $\infty$ " isn't a number so secant is not defined for it. You could, of course, extend the real number system to include "infinities" but there are several different ways to do that- you would have to specify which you meant.

But, we can talk about this in the limit sense. $sec(\theta)= \frac{1}{cos(\theta)}$ and as $\theta$ goes to $\pi/2$ , $cos(\theta)$ goes to 0 and so $sec(\theta)= \frac{1}{cos(\theta)$ "goes to infinity" (get larger without bound). Thus, if we extend the real numbers sytem in that way, we have that " $sec(\pi/2)= \infty$ " and so " $arcsec(\infty)= \pi/2$ ".

Thread: question on arcsec infinity

LinkBack

Thread Tools

Display

question on arcsec infinity

Similar Math Help Forum Discussions

If the derivative is bigger than positive constant, is the limit at infinity infinity

derivative of arcsec

arcsec proplem pls

Integration by parts of arcsec(x)

Question about derivatives of arcsec and arccsc

Search Tags