Proof of the Stolz theorem of "reprint"

Last week I wanted to write a result that has been a procrastination disorder. Last week because I do not understand the first half of the Stolz theorem proved to put the dog and put Baidu, the results of a search to find a Baidu inside the first half of the paragraph, although the answer is probably because of how to say the first half of the people who ask questions do not understand lazy to say that the latter half of the end, the result of the people very angry But I just figured out where I didn't understand. I will put two pieces together to send out, I hope that the future as I do not understand when the search can be searched here.
PS: Fihkingorts's discussion of 35 cases of 6 is too divine ... It's too much trouble ... It's horrible ... Sympathy and admiration for the terrible process that a man of theoretical mathematics has to undergo at the undergraduate stage. Sure enough, I did not go to maths but to study engineering.

(PS2: Mathematical symbols too troublesome blog and no formula function I also do not bother to use the picture, the following infinity refers to the 8,xn Yn Xn Yn N, N, n-1, N-1, etc. are subscript,/is the numerator denominator in the middle of the horizontal line, If it's hard to see the kind of formatting you write on your manuscript, it's easy to read and write. This is why I presentation and oral defense like the blackboard do not like to use PPT or anything)

First of all, the Stolz theorem, the people who will be searched here must know:
Set the whole sequence variable yn->+infinity, and from an item Yn+1>yn, then Lim (Xn/yn) =lim ((xn-xn-1)/(yn-yn-1)) (the limit to the right of the equation is known to exist)
If the limit to the right of the equation does not exist, you cannot use this theorem to kiss.

Prove:
SetLim ((xn-xn-1)/(yn-yn-1)) =a, we will proveLim (Xn/yn) is also a.

First of all, we review the whole sequence variable (high number of postgraduate examination is called series, strictly speaking, the subscript can only be an integer sequence, that is, the whole sequence variable) limit definition, the whole sequence variable xn has the definition of limit A is:
For any positive m, you can always find subscript n, making n>=n, | Xn-a|<m
When we discuss the definition of what the pro-domain AH what approaching ah all the mother threw away, are affecting the understanding of the goods.

According to the definition of the limit, for any small amount of e>0, must be able to obtain the ordinal n, so that n>=n when there is
| (xn-xn-1)/(yn-yn-1)-a|< (E/2)
(That is, the domestic high number of what this thing falls in the pro-domain of a, forget the argument, this is to make people more difficult to understand the limit)
Xn-xn-1 or infinity is hard to do, we have to find a way to solve the problem more easily.
From N to n are satisfied with the above-
| (XN-XN-1)/(YN-YN-1)-a|< (E/2)
| (Xn+1-XN)/(Yn+1-YN)-a|< (E/2)
...
| (xn-xn-1)/(yn-yn-1)-a|< (E/2)
Expand
a-e/2<(XN-XN-1)/(YN-YN-1) &LT;A+E/2
...
a-e/2< (XN-xn-1)/(YN-yn-1)<a+e/2
Because of the yn>yn-1, it is possible to have the same multiply denominator on both sides
(A-E/2)(YN-YN-1) <XN-XN-1< (A+E/2) (YN-YN-1)
(A-E/2)(Yn+1-YN)<Xn+1-XN< (A+E/2)(Yn+1-YN)
...
(A-E/2)(YN-Yn-1)<XN-xn-1< (A+E/2)(YN-yn-1)
Observation on the formula, you can find that they add up, the previous formula minus the front of the key can be eliminated after a minus sign of the item, you can get: (note n is larger than N)
(A-E/2)(YN-YN)<XN-xn< (A+E/2) n -yn That is
| ( XN-XN)/(Yn-yn)-A|<E/2
Xn and Yn are fixed numbers, which is easier to use.

Next describes an identity:
xn/yn-a= (Xn-ayn)/yn+ (1-yn/yn) ((XN-XN)/(Yn-yn)-a)
This equation is set up and can be easily checked directly. The original Stolz how to come up with this equation, too good.

Because of Yn>=yn, so 1-yn/yn <=1 is then available:
| xn/yn-a|<=| (Xn-ayn)/yn|+| (XN-XN)/(Yn-yn)-a|
when N>=n ' >n,
because Xn-ayn is a fixed number, yn->infinity, so | ( Xn-ayn)/yn|->0, with   | ( Xn-ayn)/YN|&NBSP;<E/2
At the same time has just proved n>=n | ( XN-XN)/(Yn-yn)-A|&NBSP;<E/2
therefore
| Xn/yn-a|<e
is

The proof of the Stolz theorem of "reprint"