(i) Given: limx→1(x3−x2+1)
Substituing x=1
=(1)3−(1)2+1
=1−1+1
=1
∴limx→1[x3−x2+1]=1
(ii) Given: limx→3[x(x+1)]
Substituing x=3
=3(3+1)
=3(4)
=12
∴limx→3[x(x+1)]=12
(iii) Given: limx→−1(1+x+x2+...+x10)
Substituing x=−1
=1+(−1)+(−1)2+(−1)3+(−1)4+(−1)5+(−1)6+(−1)7+(−1)8+(−1)9+(−1)10
=1+(−1+1)+(−1+1)+(−1+1)+(−1+1)+(−1+1)
=1+0+0+0+0+0
=1
∴limx→−1(1+x+x2+...+x10)=1