> 数学 >
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
人气:262 ℃ 时间:2020-04-18 12:07:00
解答
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.
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版