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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Feb 2008 20:06:42 IST
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find: 
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12 Feb 2008 20:10:01 IST
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sir but we have to know the latex language to use it and i dont know it thats the problem
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12 Feb 2008 22:09:43 IST
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I think answer is 0.i applied lhospital rule so just differentiating the numerator i get ans=0 pls correct me if wrong
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12 Feb 2008 22:10:52 IST
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am i right or not?
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12 Feb 2008 22:14:04 IST
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hey i am getting -4.
whats the ans. any ways... ??
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12 Feb 2008 22:36:25 IST
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iam gettin 0
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12 Feb 2008 22:44:04 IST
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answer is zero. But what is keeping you guys from posting the method? Mebbe there is something to be learnt here (a big hint fellas)
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12 Feb 2008 22:50:03 IST
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hey sorry guys ,
i am also getting 0 as the answer........
sorrry i got my mistake... .....
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12 Feb 2008 22:52:52 IST
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again no method. cmon guys dont be afraid to stick your neck out!
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12 Feb 2008 22:54:30 IST
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hey most of here hav done the sum with the help of l' hospital method...
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12 Feb 2008 22:56:19 IST
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lim (2cosx-2 2+cosx)/root(2+cosx) *(x-pie) x-pie Apply Lhospital rule [-2sinx + sinx/root(2+cosx) ] / (x-pie)/2root(2+cosx) + root(2+cosx) put x=pie u get 0 in numerator and 1 in denominator
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12 Feb 2008 22:56:59 IST
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if there is any other method then please post it here.
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12 Feb 2008 22:57:44 IST
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exactly,:: gettin 0 in the numerator and 1 in the denominator..
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13 Feb 2008 02:13:08 IST
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Raulrag Though you have got thre right ans,you are wrong. [ ]4cos^2x is mod[2cosx] n not 2 cosx
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13 Feb 2008 03:29:22 IST
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Re:Limit Hi guyz I have solved it but img isnt clear. So downld it and magnify 2 view. Ok guyz, Ive pasted it as 2 images in the next page, go to the next page
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