Question

求下列极限，大一微积分 ，求解，急,要有详解过程 我打勾的题目就可以了

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Answer 1

以下趋于多少均省略，注意加上：
（5）lim2^n*sinx/2^n
＝lim x*sin(x/2^n)/[x/2^n]
=xlimsin(x/2^n)/(x/2^n)
=x*1=x
(7　)lim[sqr(1-cosx)]/x =lim根号2 *sin(x/2) /x
＝sqr2 *limsin(x/2)/(x/2) /2
=(根号2）/2

（1）lim(1+3/n)^(-n)
=lim[(1+1/(n/3))^(n/3)]^(-3)
=e^(-3)
(3) lim [(x+3)/(x+1)]^x
=lim[1+2/(x+1)]^x
=lim{[1+1/((x+1)/2)]^[(x+1)/2]}^2 / lim[1+2/(x+1)]
=e^2 /1
=e^2
(5)lim(1-2x)^(1/x)
=lim{{1+1/(-1/(2x)]}^(-1/(2x))}^(-2)
=e^(-2)
(7) lim(1+tanx)^cotx
=lim(1+1/cotx)^cotx
=e

(1)limxsin(1/x)
=limsin(1/x)/(1/x)
=1
(2)limsin(ax)/sin(bx)
分子分母同时趋于0，用诺贝塔法则，分子分母同时求导数：
＝limacos(ax)/(bcos(bx))
=a*1/(b*1)
=a/b

Answer 2

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