Limit of sequence in which each term is defined by the average of preceding two terms

We have a sequence of numbers xn determined by the equality

xn=xn1+xn22

The first and zeroth term are x1 and x0.The following limit must be expressed in terms of x0 and x1
limnxn

The options are:

A)x0+2x13
B)2x0+2x13

C)2x0+3x13
D)2x03x13

Since it was a multiple choice exam I plugged x0=1 and x1=1. Which means that all terms of this sequence is 1,i.e, xn=1,nN

From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence’s limit?

Answer

2xn=xn1+xn2

2x2=x1+x02x3=x2+x12x4=x3+x22x5=x4+x3...2xn=xn1+xn2

Now sum every equation and get

2xn+xn1=2x1+x0

Supposing that xn has a limit L then making n we get:

2L+L=2x1+x0L=2x1+x03

Attribution
Source : Link , Question Author : Ananth Kamath , Answer Author : Arnaldo

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