问个极限方面的问题,在线等lim (sin(ln(x+1))-sin(ln(x)))(x->无限大）是多少?ln(x+1
2个回答

sin(ln(x+1))-sin(ln(x))

=2·cos[(ln(x+1)+ln(x))/2]·sin[(ln(x+1)-ln(x))/2]

=2·cos[(ln(x+1)+ln(x))/2]·sin[(ln(1 + 1/x)/2]

|cos[(ln(x+1)+ln(x))/2]| ≤ 1

sin[(ln(1 + 1/x)/2] → sin[ln(1)/2] = sin(0) = 0 (x → ∞)

所以

sin(ln(x+1))-sin(ln(x)) → 0 (x → ∞)

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