same can be done using L'Hospital rule (if on putting limit you get a indeterminate form like 0/0 or infinity/infinity, then keep on differentiating both numerator and denominator until you get a fininte limit)
A=lim (e^x-1)/[ln(x+1)]
x->0
on putting x=0 we get
A=(1-1)/ln(1) =0/0
so differentiating both numerator and denominator we get
A=lim (e^x)(x+1)
x->0
now putting x=0
we get A=(e^0)1=1
Since this question is posted in the PRE-Calculus subforum, I assume a solution that doesn't use calculus is required ....?
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