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Question Number 203822 by 073 last updated on 29/Jan/24
Answered by esmaeil last updated on 29/Jan/24
1:1x=p→x→∞→p→0[x+1x]=[p+1)=1→Ω=limp→0sinpp=1
2:Ω=limx→1−2x(1−x)1−x=23:Ω=limx→0+[x]−xx=limx→00−xx=−1
Answered by MathematicalUser2357 last updated on 30/Jan/24
Whatis:[[x]]
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