False. Set . For ,Iteration gives a growing geometric sequence as a lower bound for . Hence and no eventual constant can bound , as Big O notation would require.
True. Rationalizing the square root gives, for ,The uniform bound is exactly the assertion in Big O notation.
True. The logarithm identity givesFor , , so . By monotonicity of the natural logarithm, the expression lies between and . It is therefore in Big O notation.
False. On every odd positive integer , . Given any proposed Big O notation constant and threshold , choose an odd . Then , contradicting the required bound. The behavior on even integers cannot repair this failure on an unbounded subsequence.
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