00:01
For this problem, it might help to rewrite this as n minus square root of in parentheses, n plus 1 times n plus 3.
00:12
And i say that it might help to write it like this because once you just have one square root function happening here, it might remind you that, you know, we have this trick where we multiply by the top and bottom by the conjugate if we are trying to figure out what this limit is.
00:29
Okay.
00:30
And that's exactly the trick that we'll want to employ here.
00:35
We take our expression, right? think about it as a fraction, so just divide it by one.
00:45
And then we multiply the top and the bottom by the conjugate of this guy.
00:52
Okay, so we can just leave the in alone and then change the sign from negative to positive here.
01:06
And certainly if we want this to be an equality, then if we multiply by something, we have to divide by the same thing.
01:22
Okay, and now once you work all of this out, the top part we have n times n.
01:27
So we have an n squared.
01:32
We have n times the square root thing.
01:34
And then we also have the minus square root thing times n.
01:37
So the mixed terms are going to cancel out.
01:39
And we'll just have this n squared minus n plus 1 times n plus 3.
01:48
Okay.
01:49
So that's just, that's a typical thing that happens when you multiply by the conjugate.
01:53
You just get this thing squared times this thing squared.
01:59
Now we can't forget that we're still dividing by this thing over here.
02:16
Okay, and now it might help if we distribute this stuff that we have over here.
02:22
So this is n squared minus.
02:24
And now if we want to distribute this, we have n times n, so that gives us an n squared.
02:30
We have n times three.
02:32
We also have n times one, so plus 4n.
02:36
And then we have 1 times 3.
02:49
Okay, so these n squared are going to cancel.
02:56
Now we have minus 4n.
02:59
And this minus sign, remember this minus sign distributes.
03:02
As well.
03:02
So we have minus 4n minus 3 divided by n plus square root of n plus 1 times n plus 3.
03:18
And here we do the standard trick where we look in the denominator and we think about what term is going to infinity the fastest.
03:26
And then we divide the top and the bottom by whatever that term may be.
03:30
So here it looks like we have an n times n term happening.
03:34
But remember, we're taking the square root of n point...