Determine whether each of the functions log (n + 1) and log (n2 + 1) is o (log n)

Assuming the order required is as n-> inf.

As n->inf, o (log (n+1)) - > o (log (n)) since the 1 is insignificant compared with n.

We can similarly drop the "1" as n-> inf, the expression becomes log (n^2+1) - >

log (n^2) = 2log (n) which is still o (log (n)).

So yes, both are o (log (n)).

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