00:01
We are given the first term, occurrence relation, for a sequence which we are told converges, and we are asked to find its limit.
00:13
The first term is negative 4, and the recurrence relation is am plus 1 equals the square root of 8 plus 2 am.
00:25
Since the limit exists, it's some nil number l.
00:33
It's the limit of the sequence, which is the same as the limit as then it puts infinity on the square root of p plus 2 .2.
00:44
And you have that square root.
00:51
Bex is continuous from the positive real numbers.
01:06
And you have that since, so in order to prove this limit true, we just show that, so you have it, since a1 is negative 4, follows that a2 is the square root of 8 times 4, which is 0.
01:41
This is going to be in that interval in which square is continuous.
01:53
You see that the limit is increasing, and so it follows that by the continuous function beyond for limits, this is equal to the square root, the limit as n -fitches infinity, 8 plus 2 a m.
02:14
And by limit rules, this is the same as the square root of 8 plus 2 times the limit as n -fitch's infinity on a n which is simply the squared 8 plus 2 l...