WEBVTT
1
00:00:00.040 --> 00:00:02.710 A:middle L:90%
okay for this problem when our graphing it we just
2
00:00:02.710 --> 00:00:05.179 A:middle L:90%
want a plug in different values of n assuming that
3
00:00:05.179 --> 00:00:07.730 A:middle L:90%
we're just doing this by hand. So for equals
4
00:00:07.730 --> 00:00:10.630 A:middle L:90%
one, we just have three plus five, which
5
00:00:10.630 --> 00:00:26.030 A:middle L:90%
is eight case when in his one we're up here
6
00:00:26.030 --> 00:00:28.059 A:middle L:90%
at eight. And then if you want to get
7
00:00:28.059 --> 00:00:29.829 A:middle L:90%
a better graph, then you just plug in more
8
00:00:29.829 --> 00:00:33.810 A:middle L:90%
points. And when you do this, your graph
9
00:00:33.869 --> 00:00:39.799 A:middle L:90%
is going to start to look like this. You
10
00:00:39.799 --> 00:00:42.259 A:middle L:90%
see that? It looks like we're we have a
11
00:00:42.259 --> 00:00:46.649 A:middle L:90%
horizontal Assam Toto here at five. So it looks
12
00:00:49.439 --> 00:00:57.909 A:middle L:90%
. Looks like we converge two five is Remember,
13
00:00:57.909 --> 00:01:00.810 A:middle L:90%
when you're trying to figure out exactly what your sequence
14
00:01:00.810 --> 00:01:03.070 A:middle L:90%
converges to by looking at a Graff, you want
15
00:01:03.070 --> 00:01:04.109 A:middle L:90%
to keep an eye out for these horizontal aston totes
16
00:01:04.120 --> 00:01:06.909 A:middle L:90%
. So as you go further to the right,
17
00:01:06.920 --> 00:01:08.150 A:middle L:90%
what does it look like? Your functions approaching.
18
00:01:08.230 --> 00:01:10.769 A:middle L:90%
So as we go further to the right here,
19
00:01:10.769 --> 00:01:12.109 A:middle L:90%
it looks as though our function is approaching five.
20
00:01:15.340 --> 00:01:15.849 A:middle L:90%
Okay, so if we wanted to prove this,
21
00:01:23.140 --> 00:01:25.180 A:middle L:90%
what we could do is we could use the use
22
00:01:25.180 --> 00:01:27.159 A:middle L:90%
the trick where we look for the largest term that's
23
00:01:27.159 --> 00:01:30.170 A:middle L:90%
happening by largest, we mean the thing that's going
24
00:01:30.170 --> 00:01:32.750 A:middle L:90%
to infinity the fastest. So it appeared that would
25
00:01:32.750 --> 00:01:36.849 A:middle L:90%
be five to the end. So look at the
26
00:01:36.849 --> 00:01:40.159 A:middle L:90%
thing that's going to infinity, the fastest. And
27
00:01:40.159 --> 00:01:42.290 A:middle L:90%
then we can factor that out. So pull out
28
00:01:42.290 --> 00:01:44.950 A:middle L:90%
this five to the end and then we'd have ah
29
00:01:46.540 --> 00:01:52.359 A:middle L:90%
, three over five to the end here, plus
30
00:01:52.469 --> 00:01:56.810 A:middle L:90%
one right. And we're taking the square rule or
31
00:01:56.810 --> 00:02:00.269 A:middle L:90%
the end through that's multiplication function. And by that
32
00:02:00.269 --> 00:02:07.518 A:middle L:90%
we mean that it's we can write it as the
33
00:02:07.527 --> 00:02:12.467 A:middle L:90%
in through tw of five to the end, multiplied
34
00:02:12.467 --> 00:02:16.117 A:middle L:90%
by the in through tw of you know, the
35
00:02:16.117 --> 00:02:20.198 A:middle L:90%
other stuff three over five to the end, plus
36
00:02:20.198 --> 00:02:29.587 A:middle L:90%
one. So the in through is really just taking
37
00:02:29.587 --> 00:02:31.448 A:middle L:90%
whatever's in here to the one over in tower.
38
00:02:31.598 --> 00:02:34.608 A:middle L:90%
So if we take five to the end to the
39
00:02:34.608 --> 00:02:38.117 A:middle L:90%
one over in power, you should get five,
40
00:02:38.108 --> 00:02:46.478 A:middle L:90%
okay. And then over here we can pull this
41
00:02:46.478 --> 00:02:51.098 A:middle L:90%
limit inside. And once we do that, this
42
00:02:51.098 --> 00:02:53.818 A:middle L:90%
term is going to go to zero because three or
43
00:02:53.818 --> 00:03:00.527 A:middle L:90%
five others less than one an absolute value. OK
44
00:03:00.527 --> 00:03:01.358 A:middle L:90%
, so like we mentioned, this term goes a
45
00:03:01.358 --> 00:03:06.467 A:middle L:90%
zero and then we start this one here. So
46
00:03:06.518 --> 00:03:12.897 A:middle L:90%
I guess I should still have this limit. So
47
00:03:12.897 --> 00:03:15.217 A:middle L:90%
it turned out to be five times the in through
48
00:03:15.217 --> 00:03:17.288 A:middle L:90%
have won. But any root of one is still
49
00:03:17.288 --> 00:03:21.848 A:middle L:90%
going to be one. So it all simplifies to
50
00:03:21.848 --> 00:03:35.177 A:middle L:90%
just five. So the sequence converges two five.