Question

lim{(2+3x)/(3+3x)}x次方,x趋于无穷,求极限,麻烦写过程,？

Answer 1

原式=lim｛1+（1/-(3+3x）)｝^{(-(3+3x))*[x/(3+3x)]}=lime^(-x/(3+3x))=lime^(-1/(3+3/x))=e^(-1/3)
"^"表示次方 ,,7,呜呜，，，倒数第3步怎么得到倒数第二步的？谢谢。,lim{(2+3x)/(3+3x)}x次方
=lim{(3+3x-1)/(3+3x)}x次方
=lim(1-1/(3x+3))x次方
=e^lim(-x/(3x+3))
=e^(-1/3),2,当x趋于无穷时，(2+3x)/(3+3x)则趋于1，{(2+3x)/(3+3x)}x次方相当于1的x次方，则为1.,2,lim(x->无穷){(2+3x)/(3+3x)}^(x)
= lim(x->无穷) [1-1/[3+3x)]^(x)
let y = 3+3x
lim(x->无穷) [1-1/[3+3x)]^(x)
=lim(y->无穷) [1-1/[y)]^((y-3)/3)
=lim(y->无穷) [1-1/[y)]^((y/3) .lim(y->无穷) [1-1/[y)]^(-1)
= e^(-1/3). 1
=e^(-1/3),1,