Question

求极限解题过程

Answer 1

解：5.原式=lim(x->+∞){[√(x+√(x+√x))-√x][√(x+√(x+√x))+√x]/[√(x+√(x+√x))+√x]}
=lim(x->+∞){[(x+√(x+√x))-x]/[√(x+√(x+√x))+√x]}
=lim(x->+∞){√(x+√x)/[√(x+√(x+√x))+√x]}
=lim(x->+∞){√(1+1/√x)/[√(1+√(1/x+1/√x³))+1]} (分子分母同除√x)
=√(1+0)/[√(1+√(0+0))+1]
=1/2；
6. 当m=n时，原式=lim(x->1)[(x^m-x^m)/(x^m+x^m-2)]
=lim(x->1)[0/(2x^m-2)]
=lim(x->1)(0)
=0
当m=-n时，原式=lim(x->1){[x^m-x^(-m)]/[x^m+x^(-m)-2)]}
=lim(x->1){[x^(2m)-1]/[x^(2m)+1-2x^m)]} (分子分母同乘x^m)
=lim(x->1){[(x^m-1)(x^m+1)]/(x^m-1)²}
=lim(x->1)[(x^m+1)/(x^m-1)]
=∞
当m≠±n时，原式=lim(x->1){[mx^(m-1)-nx^(n-1)]/[mx^(m-1)+nx^(n-1)]}
(0/0型极限，应用罗比达法则)
=lim(x->1)[(m-n)/(m+n)]。