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分析分别解不等式，结合集合的包含关系判断即可．
解答解：由$\frac{1}{x}≥1$，解得：0＜x≤1，
由2x-1≤1，解得：x≤1，
故“$\frac{1}{x}≥1$”是“2x-1≤1”成立的充分不必要条件，
故选：A．
点评本题考查了充分必要条件，考查解不等式问题，是一道基础题．
