2305Mayb this might be of some help to you:
http://math.stackexchange.com/questions/316682/limit-of-y-sqrt5-sqrt5-sqrt5-sqrt5-sqrt5-ldots/316691#316691
8 Answers
1357y2=5+x
x2=5-y
_____________________
y2-x2=x+y
y-x=1
y2=5+y-1=4+y
y2-y=4
Sayan Sinha Thank You sir. The solution is so simple and I was going on and on doing this and that. Now I understand how valuable your hint was. Thanx once again...Upvote·0· Reply ·2013-04-27 04:26:07
1357let x=√5-√5+√5....
then we have
y=√5+x
and
x=√5-y
proceed from here to get the answer
Welcome in Advance :P
111
- Sayan Sinha No...right answer is (3) 4....but I can't understand how...
Thanx...
Sayantan Hazra 1 is not the right answer....
@sayan, Manish sir has already given you the hint....after obtaining a relation between x and y, evaluate y^2 - y.
2305337Yeah! It does answer my question but it leaves me with one doubt. It's written over there in the first line:
It is also clear that y2≥5 and 0<y.
Why is it so?
Thanx...
2305Square root of any number is positive.
→y>0
y^{2} = 5+\sqrt{5-\sqrt{5+\sqrt{5-\dots}}}→ y^{2}\geq 5 as
\sqrt{5-\sqrt{5+\sqrt{5-\dots}}}>0
- Shaswata Roy Sorry for the last line,guess there was some kind of problem. Anyways I hope that you get my point.
337Square root of any number is ± of another number. It's not necessary it has to be positive.
Please can you tell me where I am going wrong...?
Thanx again...
- Sayantan Hazra it is quite clear from the question that this is the positive square root, as y>0 (look at the ques: the expression of y), and directly from this, we can infer that y^2 >0.