lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 03:00:39
lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?这属於1

lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?
lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?

lim(x->无穷)[(5x+1)/(5x-1)]^(4x)=?
这属於1^(无穷) 类形,主要都是用自然对数然後用洛必特法
设 y= [5x+1)/(5x-1)]^(4x)
lny = ln{[5x+1)/(5x-1)]^(4x)}
lny = ln{[5x+1)/(5x-1)]^(4x)}
= 4x ln{[5x+1)/(5x-1)}
= 4 ln{[5x+1)/(5x-1)} / (1/x) 当 lim(x->无穷 ) 属於0/0 的情况
所以可以用 洛必特法
分子分母各取导数得
lim(x->无穷)ln y =lim(x->无穷) 4 ln{[5x+1)/(5x-1)} / (1/x)
=lim(x->无穷) 4 {-10/(5x+1)(5x-1)} /(-1/x^2)
=lim(x->无穷)40 [ x^2/((5x+1)(5x-1))]
lim(x->无穷)ln y = 40/25= 8/5
所以lim(x->无穷)y= e^(8/5)

定义y=5x-1
[(5x+1)/(5x-1)]^(4x)=(1+2/y)^((y+1)/5*4)
=((1+2/y)^(y/2))^{(y+1)/y*8/5}
-->e^(8/5)

=lim(x趋于无穷)[(5x-1+2)/(5x-1)]^4x
=lim(x趋于无穷)[1+2/(5x)]^[(5x/2)(8/5)] 这个形式知道吧
=e^8/5