Question

求高数大神

Answer 1

四.1. lim(x->1)(1/(1-x)-n/(1-x^n))
=lim(x->1)((1+x+x^2+…x^(n-2)+x^(n-1)-n)/(1-x^n))
用罗必达法则
=lim(x->1)((1+2x+...+(n-2)x^(n-3)+(n-1)x^(n-2))/(-nx^(n-1)))
=(1+2+...+(n-2)+(n-1))/(-n)=-(n-1)n/2/n=(1-n)/2
2.
1）当m=n时，lim(x->1)(m/(1-x^m)-n/(1-x^n))=0
2）当m≠n时，
lim(x->1+)(m/(1-x^m)-n/(1-x^n))
=lim(x->1+)((m-mx^n-n+nx^m)/((1-x^m)(1-x^n)))
lim(x->1+)((m-mx^n-n+nx^m)/(1-x^m-x^n+x^(m+n)))
用罗必达法则
=lim(x->1+)((-mnx^(n-1)+mnx^(m-1))/(-mx^(m-1)-nx^(n-1)
+(m+n)x^(m+n-1)))=0 （因分母的幂次高于分子）
当|x|<1时，设x=1/t, 则
lim(x->1-)(m/(1-x^m)-n/(1-x^n))
=lim(t->1+)(m/(1-（1/t)^m)-n/(1-(1/t)^n))