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 Discussion Response Post to:
Limits
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Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2008 12:03:13 IST
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as x tends to 1...numerator approaches 0....but denomenator is 1+a+b!
since the limit exists...denominator should also approach 0...
=> 1+a+b =0
we apply LH rule...
-2ln(2-x) /(2-x)(2x+a)
as x approaches 1...numerator approaches 0...so den. should also approach 0....s0 (2-1)(2+a) = 0
=> a=-2
and b = 1
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this reply: 5 points
(with 1 
in 1 votes ) [?]
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