Proof of limit

Find lim (x trends infinity) x^(1/x)

ln x ^(1/x) = ln x / x

If lim (x trends infinity) x^(1/x) exist if and only if lim (x trends infinity) ln x ^(1/x) exist.


Let y = ln x, as x trends infinity, y trends infinity, we have lim (x trends infinity) ln x ^(1/x) = lim (y trends infinity) y / exp(y)

as exp(y) = 1 + y + y^2/2 + .. + y^n / n! + ..... for all y

lim (y trends infinity) y / exp(y) = lim (y trends infinity) y / (1 + y + y^2/2 +....)
= 0, as y is about to equals 1+y for a large y, y^2/2 + ... with be much much larger.

We have lim (x trends infinity) ln x ^(1/x) = 0

We have lim (x trends infinity) x^(1/x) = exp(0) = 1

2008-10-20 13:13:33 補充：
In fact, I used L'hospital rule indirectly. As all function can be written in
f(x+dx)=f(x) + f'(x)dx + f''(x)dx^2 +..... for all f(y) exist at y=x

2008-10-20 13:16:01 補充：
So it is easy to see I expand exp(y) into that form, that is if I do some rewrite, you can have the first term as zero, it will become L'hospital rule.

2008-10-20 13:19:03 補充：
g(x+h)/f(x+h) = [g(x) + g'(x)h + ...]/[f(x)+f'(x)h+.....], if g(x) = f(x) = 0, it becomes g(x+h)/f(x+h) = [ g'(x)h + ...]/[f'(x)h+.....], the L' hospital rule.

I like kachun.charles's proof.
But there is a catch: without knowing a yet, we dunno if |a| <1 or not, so strictly speaking we can't apply the binomial theorem directly.
One remedy is to let x be integers -> infinity, and use the fact that x^(1/x) is continuous for x>0.

Let a=x^(1/x)-1
then x=(1 a)^x=1 xa a^2x(x-1)/2 ...>1 a^2x(x-1)/2
2/x>a^2
for every e>0 , Take N=2/e^2 1 ,
s.t. |a-0|<sqrt(2/x)<e for all n>N
i.e. result follows

2008-10-20 13:40:15 補充：
Or you can apply Sandwich Thm~