Find LIM (n → + ∞) (1 + n) [ln (1 + n) - ln n]

Find LIM (n → + ∞) (1 + n) [ln (1 + n) - ln n]

Find LIM (n →∞) ln (n!) / ln (n ^ n)

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lim（n->∞）ln(1+1/x)/arccotx?

In fact, there is no limit. When x tends to positive infinity, it belongs to 0 / 0 type. First, ln (1 + 1 / x) is equivalent to 1 / X by using the equivalent infinitesimal when x tends to positive infinity, and then lobita's law can be used to obtain: the original formula = LIM (x → + ∞) [(- 1 / x ^ 2) / [- 1 / (1 + x ^ 2)] = LIM (x → + ∞) [(1 + x ^ 2) / (x + x ^ 2)] = L