题目内容：

求函数极限 lim(x->1)(4/(1-x^4)-3/(1-x^3)) lim(x->无穷大）（x^7(1-2x)^8/(3x+2)^15）
lim(x->1)((x^(n+1)-(n+1)x+n)/(x-1)^2)
lim(x->1)(4/(1-x^4)-3/(1-x^3))
lim(x->无穷大)(x^7(1-2x)^8/(3x+2)^15）

优质解答

（1）lim(x->1)[4/(1-x^4)-3/(1-x³)]=lim(x->1){4/[(1-x)(1+x)(1+x²)]-3/[(1-x)(1+x+x²)]}
=lim(x->1){[4(1+x+x²)-3(1+x)(1+x²)]/[(1-x)(1+x)(1+x²)(1+x+x²)]}
=lim(x->1){(1+x+x²-3x³)/[(1-x)(1+x)(1+x²)(1+x+x²)]}
=lim(x->1){(1-x)(1+2x+3x²)/[(1-x)(1+x)(1+x²)(1+x+x²)]}
=lim(x->1){(1+2x+3x²)/[(1+x)(1+x²)(1+x+x²)]}
=(1+2+3)/[(1+1)(1+1)(1+1+1)]
=1/2；
（2）lim(x->∞)[x^7(1-2x)^8/(3x+2)^15]=lim(x->∞)[(1/x-2)^8/(3+2/x)^15]
=(0-2)^8/(3+0)^15
=2^8/3^15；
（3）lim(x->1){[(x^(n+1)-(n+1)x+n]/(x-1)²}=lim(x->1){[(n+1)x^n-(n+1)]/[2(x-1)]}
(0/0型极限,应用罗比达法则)
=lim(x->1){[n(n+1)x^(n-1)]/2}
(0/0型极限,再次应用罗比达法则)
=n(n+1)/2.