Q7. limy→1y−1y3−1
\section*{Solution:} y→1limy−1y3−1=y→1limy−1(y)3−(1)3=y→1limy(y−1)(y2+y+1) =y→1lim(y2+y+1)=(1)2+1+1=1+1+1=3
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