December 5th, 2010, 10:44 AM  #1 
Newbie Joined: Dec 2010 Posts: 5 Thanks: 0  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 
December 5th, 2010, 11:02 AM  #2 
Senior Member Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0  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 . 
December 5th, 2010, 11:07 AM  #3 
Newbie Joined: Dec 2010 Posts: 5 Thanks: 0  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 
December 5th, 2010, 11:14 AM  #4 
Senior Member Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0  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.

December 5th, 2010, 11:18 AM  #5 
Newbie Joined: Dec 2010 Posts: 5 Thanks: 0  Re: limit (cos(1\x))tan x x> .5pi
Thats it? pretty simple! thanks 
December 5th, 2010, 11:26 AM  #6 
Senior Member Joined: Nov 2010 Posts: 288 Thanks: 1  Re: limit (cos(1\x))tan x x> .5pi
hello there the limit is plus infinity 

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5pi, cos1xtan, limit, x> 
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