Question

求解这个极限（求详解）

网上好多种答案也不知道哪一个了，求高数大佬的解答比心♥️

Answer 1

你可以乘上根号（n+2）加上根号n，再除以他，配出个平方差，上面就是2倍根号n，下面其实也是，答案就是1了吧

Answer 2

lim(x->0) [(1+x)^x -1 ]/ lncosx
=lim(x->0) [ (1+x^2.(1/x) )^x -1 ]/ lncosx
=lim(x->0) [ e^(x^2) -1 ]/ lncosx
=lim(x->0) x^2/[ (-1/2)x^2]
=-2
OR
x->0

分母
cosx =1- (1/2)x^2 +o(x^2)
lncosx = ln[1- (1/2)x^2+o(x^2)] = -(1/2)x^2 +o(x^2)
分子
根据泰勒展式
f(x) = (1+x)^x => f(0) =1
f(x) =(1+x)^x

lnf(x) = xln(1+x)

(1/f(x)) f'(x) = [x/(1+x) + ln(1+x) ]
f'(x) = [x/(1+x) + ln(1+x) ] . (1+x)^x
=> f'(0) =0
f''(x) ={ 1/(1+x)^2 + 1/(1+x) +[x/(1+x) + ln(1+x) ]^2 } . (1+x)^x
=> f''(0) = 2
(1+x)^x = 1 + x^2 +o(x^2)
(1+x)^x -1 = x^2 +o(x^2)
//
lim(x->0) [(1+x)^x -1 ]/ lncosx
=lim(x->0) x^2/ [ -(1/2)x^2 ]

=-2