Limit Problem

Let L＝lim(1＋1/r)^r
　　　r→∞

ln L＝ln[lim(1＋1/r)^r]
　　　 r→∞
　　＝lim ln[(1＋1/r)^r]
　　　r→∞
　　＝lim r ln(1＋1/r)
　　　r→∞

Let h＝1/r

ln L＝lim ln(1＋h)/h
　　 h→0
　　＝lim[ln(1＋h)－ln(1)]/h
　　　h→0
　　＝(d/dx)(lnx) at x＝1
　　＝1/x at x＝1
　　＝1

＝＞　L＝exp(1)＝e

A rigorous proof from Principles of Mathematical Analysis:

Let s(r)＝1/1!＋1/2!＋．．．+1/r!
　 t(r)＝(1＋1/r)^r

By the binomial theorm,
t(r)＝1＋1＋(1－1/r)/2!＋(1－1/r)(1－2/r)/3!＋．．．＋(1－1/r)(1－2/r)…[1－(r－1)/n]/r!

Hence t(r)≤s(r), and

lim sup t(r)≤lim sup s(r)＝e　　　(1)
r→∞　　r→∞

Next, if r≥m,

　　　　t(r)≥1＋1＋(1－1/r)/2!＋(1－1/r)(1－2/r)/3!＋．．．＋(1－1/r)(1－2/r)…[1－(m－1)/r]/m!
＝＞　　lim inf t(r)≥1/1!＋1/2!＋．．．+1/m!＝s(m)
　　　　r→∞

Therefore,

lim inf t(r)≥lim inf s(m)＝e　　　(2)
r→∞　　m→∞

From (1) and (2), we have

lim t(r)＝e
r→∞

2008-07-02 21:25:25 補充：
Rewrite the above result as follow

lim t(n)＝e　　　　　　　　　　　　　　　(3)
n→∞

where n＝1,2,3,…

For real x＞0, we have

　　　　t(x)＝(1＋1/x)^x
＝＞　　ln t(x)＝x ln (1＋1/x)
＝＞　　t′(x)/t(x)＝－1/[x(1＋1/x)]＋ln(1＋1/x)
　　　　　　　＝－1/(x＋1)＋ln(1＋1/x)
　　　　　　　＝ln{exp[－1/(x＋1)]}＋ln(1＋1/x)
　　　　　　　＝ln{(1＋1/x)/exp[1/(x＋1)]}　　　　　(4)

2008-07-02 21:26:24 補充：
As x＞0　　＝＞　　0＜1/(1＋x)＜1

we have

1＋1/x＝1/[1－1/(x＋1)]＝1＋1/(x＋1)＋1/(x＋1)²＋1/(x＋1)³＋．．．
　　　≥1＋1/(x＋1)＋1/[2!(x＋1)²]＋1/[3!(x＋1)³]＋．．．
　　　＝exp[1/(x＋1)]

＝＞ (1＋1/x)/exp[1/(x＋1)]≥1
＝＞ ln{(1＋1/x)/exp[1/(x＋1)]}≥0

From (4), we have

t′(x)＝t(x)ln{(1＋1/x)/exp[1/(x＋1)]}≥0

＝＞　　t(x) is monotonically increasing

2008-07-02 21:27:08 補充：
For x＞0, there exists an integer n(x)＞0 such that

n(x)≤x≤n(x)＋1

and

lim n(x)＝lim n(x)＋1＝∞
x→∞　　x→∞

As t(x) is monotonically increasing, we have

t（n(x)）≤ t(x) ≤t（n(x)＋1）　　　 　　　　　　　　　　　(5)

2008-07-02 21:27:29 補充：
From (3), we have

lim t（n(x)）＝e　　　　　　　　　　　　　　(6)
x→∞

and

lim t（n(x)＋1）＝lim t（n(x)＋1）＝e　　　　　(7)
x→∞　　　　　n→∞

From (5), (6) and (7), we have

lim t(x) exists and equals to e.
x→∞

2008-07-02 21:32:07 補充：
Correct for the line "For x＞0, there exists an integer n(x)＞0 such that":

For x＞0, there exists an integer n(x)≥0 such that

2008-07-03 07:32:02 補充：
Correction again for the same line － not to consider t(0):

For x＞1, there exists an integer n(x)≥1 such that

2008-07-04 14:08:34 補充：
Correction for (7):

lim t（n(x)＋1）＝lim t（n(x)＋1）＝e　　　　　(7)
x→∞　　　　　x→∞

2008-07-05 09:32:38 補充：
Correction again－ above correction for (7) is wrong as the original version is right:

lim t（n(x)＋1）＝lim t（n(x)＋1）＝e　　　　　(7)
x→∞　　　　　n→∞

To hawk_wing_1999:
Please read carefully my question.

To ken_29_2003:
r is real number, but not only integer

To nychan63:
Your first proof is supposing that L has limit, but this question requires you to prove that L has limit
And your second proof is supposing that r is integer

hawk_wing_1999 is wrong!

用binomial expend [1+(1/r)]^r , 發覺佢收歛得好快.

so , lim(r-> infinity) [1+(1/r)]^r 會 等於一實數 .

in fact , lim(r-> infinity) [1+(1/r)]^r = e

2008-06-30 12:30:26 補充：
樓上個位做到lim(r->infinity)(1^r) 係不定式, 出唔到答案的..

lim(r->infinity)(1+1/r)r
=lim(r->infinity)[(r+1)/r]r
=lim(r->infinity)(1r) (L'Hopital Rule)
=1
∴lim(r->infinity)(1+1/r)r has limit as r is real and tends to infinity, and the limit is =1