Question

高数 求极限 详解谢谢

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Answer 1

（2）令t=x-1，则x=1+t
原式=lim(t->0) [(1+t)^m-1]/[(1+t)^n-1]
=lim(t->0) (mt)/(nt)
=m/n
（4）原式=lim(x->∞) {[1-2c/(x+c)]^[-(x+c)/2c]}^[-2cx/(x+c)]
=e^(-2c)
=9
所以c=-ln3

Answer 2

(2)
lim(x->1) (x^m-1)/(x^n-1)
=lim(x->1) (x-1)[x^(m-1)+x^(m-2)+...+1]/{ (x-1)[x^(n-1)+x^(n-2)+...+1]}
=lim(x->1) [x^(m-1)+x^(m-2)+...+1]/[x^(n-1)+x^(n-2)+...+1]
=m/n
(4)
(x-c)/(x+c)= 1- 2c/(x+c)
let
2c/(x+c) = 1/y
lim(x->∞)[ (x-c)/(x+c)]^x =9
lim(x->∞)[ 1- 2c/(x+c)]^x =9
lim(y->∞) ( 1- 1/y)^(2cy-c) =9
lim(y->∞) ( 1- 1/y)^(2cy) =9
e^(-2c) = 9
-2c=2ln3
c= -ln3