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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:(1/x)=p→^(x→∞→p→0) [((x+1)/x)]=[p+1)=1→  Ω=lim_(p→0) ((sinp)/p)=1

1:1x=pxp0[x+1x]=[p+1)=1Ω=limp0sinpp=1

Answered by esmaeil last updated on 29/Jan/24

2:Ω=lim_(x→1^− ) ((2x(1−x))/(1−x))=2  3:Ω=lim_(x→0^+ ) (([x]−x)/x)=lim_(x→0) ((0−x)/x)=−1

2:Ω=limx12x(1x)1x=23:Ω=limx0+[x]xx=limx00xx=1

Answered by MathematicalUser2357 last updated on 30/Jan/24

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