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We have to verify that lim x-->0 [ ln(1+x)/x] = 1.
substituting x = 0, we get the indeterminate form 0/0, therefore we can use the l'Hopital's rule and substitute the numerator and denominator with their derivatives.
=> lim x-->0 [ ((1/(1+x))/1]
substituting x = 0
=> 1/(1 + 0)
This verifies that lim x-->0 [ ln(1+x)/x] = 1.
In other words, we'll have to prove that:
lim [ln(1+x)]/x = 1
We'll re-write the function:
lim (1/x)*ln(1+x) = 1
We'll apply the power property of logarithms:
lim ln [(1+x)^(1/x)] = ln lim [(1+x)^(1/x)]
But, for x->0 lim [(1+x)^(1/x)] = e (remarcable limit)
lim ln [(1+x)^(1/x)] = ln e
We know that ln e = 1
So, for x->0, lim ln [(1+x)^(1/x)] = 1.
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