00:01
We want to determine what the sequence, a .n is equal to n squared over 2n minus 1 times sign of 1 or n converges or diverges, and if it converges, what it converges to.
00:13
So first, let's just look at what happens as end goes to infinity for this expression.
00:18
So as end goes to infinity, well, we know from when we first started working with limits for n behavior of functions, n squared over 2n minus 1, well since the number is larger than the denominator, that should go to infinity.
00:36
And sign of 1 over, or sign of 1 over in, well, 1 over n goes to 0, and then sign of 0 is just 0.
00:47
So now i don't know what infinity times 0 is, but this means that we could rewrite this in some way, so we can use low -pecalor rule.
01:03
To do is i'm going to reciprocate n squared over 2n minus 1, or not reciprocate, but write it as 1 2n minus 1 over n squared.
01:16
So, and then time, sine of 1 over n, and now let's divide the n squared into 2n minus 1, and doing that we'd get, so sine of 1 over n divided by 2 over n minus 1 over n squared.
01:38
And now notice as n goes to infinity for this, so once again, our numerator goes to 0, 2n, or 2 over n goes to 0, and 1 over n squared goes to 0.
01:51
So we get 0 over 0, which implies we can use lopetal's rule for this.
01:57
Now let's go ahead and look at the limit as n approaches infinity.
02:02
Okay, yep.
02:05
Well, by lopetal's rule, we know this is going to be equal to the limit.
02:12
As in approaches infinity of.
02:17
So we take the derivative of our numerator, so we're going to have to use chain rule...