题目

n平方*(2的n分之1次方-2的n+1分之1次方)在n趋于无穷时的极限

提问时间：2020-09-16

答案

令x=1/n,则x→0,
原式=limx→0 {2^x-2^[x/(x+1)}/x^2
=limx→0 2^[x/(x+1)]*{[2^[x^2/(x+1)]-1}/x^2
=limx→0 2^[x/(x+1)]*limx→0 {[2^[x^2/(x+1)]-1}/x^2
=2^0*limx→0 {[2^[x^2/(x+1)]-1}/x^2
=limx→0 {[2^[x^2/(x+1)]*ln2*[x(x+2)/(x+1)^2]}/2x
=limx→0 {[2^[x^2/(x+1)]*ln2*(x+2)/2(x+1)^2
=2^(0/1)*ln2*(0+2)/2*(0+1)^2
=ln2.