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December 5th, 2010, 09:44 AM   #1
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limit (cos(1\x))|tan x| x--> .5pi

1. having a hard time figuring out. I know the answer if infinite but having a hard time proving it

2. tan x = cot (.5pi - x) did not help

3. tried substituting t = 1\x got stuck

thanks
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December 5th, 2010, 10:02 AM   #2
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Re: limit (cos(1\x))|tan x| x--> .5pi

approaches a positive finite number as .

approaches as

So you can simply multiply the two limits. A positive finite number times is .
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December 5th, 2010, 10:07 AM   #3
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Re: limit (cos(1\x))|tan x| x--> .5pi

thanks

but i do not quiet understand what you mean

i know the squeeze the theorem but i do not understand how to use it

thanks
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December 5th, 2010, 10:14 AM   #4
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Re: limit (cos(1\x))|tan x| x--> .5pi

Never mind - you don't need the squeeze theorem for this. I'm going to edit the above post with the solution which is much simpler.
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December 5th, 2010, 10:18 AM   #5
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Re: limit (cos(1\x))|tan x| x--> .5pi

Thats it?

pretty simple!

thanks
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December 5th, 2010, 10:26 AM   #6
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Re: limit (cos(1\x))|tan x| x--> .5pi

hello there
the limit is plus infinity
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