How do we evaluate limit of,
lim x> 0 [ln(x+1)/( (2^x)  1)]
I tried using the substitution x+1 = e^k , when x tends to 0 so does k, which gave out,
lim k>0 [ k/((2^((e^k)  1)) 1 ) ]
which I simplified into( for the ease of use let e^k =a)
lim k>0 [k/( ((2^a)/2)  1)]
lim k>0 [2k/((2^a) 2 )]
which sums up to,
(2*0)/((2^(e^0))  2)
(2*0)/(22)
which gives out an undefined value as 0/0
How do we solve this?
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1 answer

How about good old L'Hopital's rule
lim x> 0 [ln(x+1)/( (2^x)  1)]
= lim x> 0 [1/(x+1) / (ln2*2^x)
= (1/1)/(ln2 ( 1)
= 1/ln2 = appr 1.44269..
check: let x = .0001 in the original
= ln(1.0001) / (2^.0001  1)
= 1.4425 , not bad, my answer is correct 👍
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