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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Hey, I looked on a graphing calculator and found that the limit of acos(x)-(pi/2) as x goes to 0 equals 0. Here's the only iffy part: is pi/2 a coincidence just popping up as some random number, or does the acos function acually really relate to pi/2?:/

Linux FTW

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

its because acos is the inverse function of the cosine function. The reason pi appears is because we're working with radians.

pi radians = 180 degree's, so pi/2 radians = 90 degrees

cos(90) = 0 so acos(0) = 90. But since we like to work with radians rather then degrees to describe an angle, the angle 90 degrees is rewritten as pi/2 radians.

So acos(0) = pi/2. :-)

Its also cool to note that the limit of tan x as x approaches pi/2 from the left (90 degree's) is infinity. Therefore the limit of atan(x) as x approaches infinity, is pi/2. :-)

A logarithm is just a misspelled algorithm.

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