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Differential Calculus
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venkat v.t
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Joined: 7 Aug 2007
Posts: 244
9 Sep 2007 22:28:24 IST
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Answer is 0.
Since in the nbd of 0 cosx lies between 0 and 1.(Since 1st or 4th quadrant cos +ve)
So [cosx]=0
So given limit is sin0/1+0 =0.
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0 sin[cos x]/1+[cos x] where [ ] denotes the greatest integer function.







