lim(x→1)f(x)＝lim(x→1)[x^3+sin(x^2-1)/(x-1)+2lim(x→1)f(x)]
＝1+lim(x→1)[sin(x^2-1)/(x-1)]+2lim(x→1)f(x)
即lim(x→1)f(x)=-1-lim(x→1)[sin(x^2-1)/(x-1)]
而lim(x→1)[sin(x^2-1)/(x-1)]
=lim(x→1){[sin(x^2-1)/(x^2-1)]•(x^2-1)/(x-1)}
=lim(x→1)[(x+1)(x-1)/(x-1)]
=lim(x→1)(x+1)=2
所以 lim(x→1)f(x)=-1-2=-3
f(x)=x^3+sin(x^2-1)/(x-1)-6lim(x→1)f(x)＝lim(x→1)[x^3+sin(x^2-1)/(x-1)+2lim(x→1)f(x)]＝1+lim(x→1)[sin(x^2-1)/(x-1)]+2lim(x→1)f(x)這步驟最后一點(diǎn)有個疑問！就是對f（x）求極限的時候，最后2lim(x→1)f(x)為什么就不對其求極限了？？？？lim(x→1)f(x)存在，是一個常數(shù)，極限等于它本身。如設(shè)lim(x→1)f(x)＝C，則lim(x→1)[lim(x→1)f(x)]＝lim(x→1)C=C=lim(x→1)f(x)