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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Nov 2007 19:58:06 IST
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[ x] [o ] (sin ax)/(sin bx)
plz give me the procedure..!!
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25 Nov 2007 20:00:27 IST
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divide both the numerator n denominator by x so u get it as a identity
the ans is a/b
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25 Nov 2007 20:00:43 IST
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arrey easy yaar... it is 0/0 form , so by lh rule the limit would be lt x-->0 acosax/bcosbx put x=0 the answer is a/b ...simple...
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25 Nov 2007 20:01:58 IST
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It has to be a/b
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25 Nov 2007 20:05:51 IST
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[ x] [o ] (sin ax)/(sin bx)
=(a/b)[ x][o ] (sin ax/ax) X (bx/sin bx)
=(a/b) (1X1)
=a/b
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25 Nov 2007 20:07:39 IST
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hi sinjan...
as u cld c its of 0/0 form...
if u apply the limit thn
limit X--->0 acosax/bcosbx
if u substitute x=0
the answer wld be a/b.
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25 Nov 2007 20:07:51 IST
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sinax/sinbx = [(sinax)/(ax)] [(bx)/(sinbx)] [ax/bx] = 1.1.a/b as x=> 0 = a/b......ans..
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25 Nov 2007 20:08:48 IST
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i found 1 more way of doing it..!!
divide both numerator and denominator by abx.
you can then easily get a/b....
thanx guys.!!
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25 Nov 2007 20:09:43 IST
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yes, both the ways r correct !
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25 Nov 2007 20:19:08 IST
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ncert question hai
use the property : limit x-> 0 sinx/x = 1
=> (sin ax)/(sin bx)
= [ ax * (sin ax) / ax ] / [ bx * (sin bx) / bx ]
= a/b *1 * 1
= a/b
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