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Nandana
110 Points
4 years ago
hi , it’s really interesting question ! you might try in this way – lim x → 0 (2x -1)/ (√(1+x) -1) = lim x → 0 (exln2 -1 ) / (1+ ½x -1 ) [ . . . ax = ex ln a & as x → 0 √(1+x) = 1+ ½ x , from Binomial expansion ] = lim x → 0 ( 1+ (x ln 2) / 1! + (x ln2 )2 / 2! + … – 1 ) / ½ x = 2 ln 2 = 1.386294361
Nandana
110 Points
4 years ago
hi ,
it’s really interesting question !
you might try in this way –
Lim x → 0  (2x -1)/ (√(1+x) -1) = Lim x → 0  (exln2 -1 ) /  (1+ ½x -1 )
[ . . .  ax = ex ln a  & as x → 0  √(1+x) = 1+ ½ x , from Binomial  expansion ]

=  Lim x → 0 ( 1+  (x ln 2) / 1! + (x ln2 )2 / 2! + … – 1 ) / ½ x
= 2 ln 2 =  1.386294361