Solve Please
1
Name:
Anonymous
2008-08-13 18:05
lim(n->INF)
2
Name:
Anonymous
2008-08-13 18:08
for neatness:
3
Name:
Apollo
2008-08-13 18:10
divide top and botton by n
4
Name:
Anonymous
2008-08-13 18:14
n^2+1
5
Name:
Anonymous
2008-08-13 18:21
You can't apply limits like that...lim(n->inf) just means there is no limit.
6
Name:
Anonymous
2008-08-13 21:07
>>5
yes you can, and the marking scheme's right.
(n^2+1)/((n+1)^2+1) = (1+1/n^2)/(1+2/n+2/n^2)
take lim(n->infinity) and you get 1 (1/n etc tend to 0)
7
Name:
Anonymous 2008-08-13 21:35
abstract nonsense.
8
Name:
Anonymous
2008-08-14 0:30
this thread needs some tex\lim_{n\to \infty}\frac{n^2+1}{(n+1)^2+1} = \lim_{n\to \infty}\frac{n^2+1}{n^2+2n+2} =\lim_{n\to \infty}1 -\frac{2n+1}{n^2+2n+2} = \lim_{n\to \infty}1 -\frac{2+1/n}{n+2+2/n} = 1 + 0 \lim_{n\to \infty}\frac{n^2+1}{(n+1)^2+1} = \lim_{n\to \infty}\frac{n^2+1}{n^2+2n+2} =\lim_{n\to \infty}\frac{1+1/n^2}{1+2/n+2/n^2} = \frac{1+0}{1+0+0}
9
Name:
Anonymous
2008-08-14 10:55
It's pretty obviously 1.
>>8
shows a pretty mathematically sound argument, but merely by observing that it's the quotient of two monic polynomials of the same degree shows it obviously tends to 1 or -1.
10
Name:
Anonymous
2008-08-15 1:33
Actually I believe the answer is 1.154
11
Name:
Anonymous
2008-08-15 5:02
since top and bottom both approach inf cant you just use L'Hopital's?
12
Name:
Anonymous
2008-08-15 8:46
how is the hospital going to help?
13
Name:
4tran 2008-08-16 4:02
>>11
Yes, you can.
>>12
Troll.
14
Name:
Anonymous
2008-08-18 1:09
15
Name:
Anonymous
2008-08-18 13:41
>>12
troll or not that made me laugh
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