Question

高数 求极限题目，第二小题。。

Answer 1

(2)对原式取自然对数得
n^2*ln[cos(x/n)]=ln[cos(x/n)]/(1/n^2)(n->∞)
=ln[cos(x/n)]/(1/n^2)(n->∞)
=[1/cos(x/n)][-sin(x/n)](-x/n^2)/(-2/n^3)(n->∞)(罗必塔法则)
=1*(-x/n)(-x/n^2)/(-2/n^3)(n->∞)
=(-x^2)/2
∴原式=e^[-(x^2)/2]=1/√[e^(x^2)]
解法2
[cos(x/n)]^(n^2)={1-2[sin(x/2n)]^2}^(n^2)(n->∞)
={1-2(x/2n)^2}^(n^2)
=[1-x^2/(2n^2)]^(n^2)
=[1-x^2/(2n^2)]^[(-2n^2/x^2)(-x^2/2)](n->∞)
=e^(-x^2/2)

Answer 2

n->无穷，x/n->0,cos(x/n) -->1,所以整体趋近于1

Answer 3

我以为你说的第四题，所以我二、四题都给写了

追答