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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Mar 2007 12:16:20 IST
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[ x] [ /2 ] (sec7x.cos3x)
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"OUR GREATEST GLORY IS NOT IN NEVER FALLING.BUT IN RAISING EVERTIME WE FALL".
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29 Mar 2007 17:23:09 IST
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Apply L'Hospital rule, since this is in 0/0 form
[x ] [  /2] [d(cosx)/dx]/d[(cos7x)/dx]
=[x ] [ /2] sinx/(7sin7x)=1/sin(4-/2)
=-1/7
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29 Mar 2007 18:22:31 IST
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hey yahiya its not cosx,its cos3x
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Don't tell me that a problem is difficult one,if it won't be difficult it would not be a problem. |
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29 Mar 2007 18:23:33 IST
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ritu is the ans 9/49
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Don't tell me that a problem is difficult one,if it won't be difficult it would not be a problem. |
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sec7x.cos3x = cos3x/cos7x
using l-hospital's rule,
we get
given limit=3/7
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29 Mar 2007 18:53:29 IST
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thanks aysh.the ans is 3/7.
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"OUR GREATEST GLORY IS NOT IN NEVER FALLING.BUT IN RAISING EVERTIME WE FALL".
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