
Title: Theorem on Limits

Series: Real Analysis

Chapter: Sequences and Limits

YouTubeTitle: Real Analysis 4  Theorem on Limits

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Subtitle on GitHub: ra04_sub_eng.srt

Timestamps
00:00 Intro
00:18 Limit of a sequence
01:16 Theorem on limits
03:48 Example
06:22 Outro

Subtitle in English
1 00:00:00,470 –> 00:00:03,170 Hello and welcome to real analysis.
2 00:00:03,770 –> 00:00:09,150 and as always many, many thanks to all the nice people that support this channel on Steady or Paypal.
3 00:00:09,780 –> 00:00:13,700 In todays part 4 we will talk about the so called limit theorems.
4 00:00:14,340 –> 00:00:17,820 Indeed, when you deal with sequences you use them all the time.
5 00:00:18,640 –> 00:00:20,040 For this please recall.
6 00:00:20,060 –> 00:00:24,350 This is how we denote a sequence which consists of real number “an”
7 00:00:25,110 –> 00:00:28,200 and if the sequence is convergent to a point “a”
8 00:00:28,450 –> 00:00:30,490 we have a notation for this limit.
9 00:00:31,180 –> 00:00:33,180 One possibility is to write
10 00:00:33,190 –> 00:00:37,180 limit “n to infinity” of “an” is equal to “a”
11 00:00:38,250 –> 00:00:41,350 and another possibility is just to use an arrow,
12 00:00:41,370 –> 00:00:45,450 where we write “n to infinity” and then it goes to “a”
13 00:00:46,370 –> 00:00:48,370 Both things exactly mean the same
14 00:00:48,380 –> 00:00:51,180 Namely the definition we gave for convergence.
15 00:00:51,760 –> 00:00:53,710 In this picture you see it again.
16 00:00:53,730 –> 00:00:59,450 For all epsilon the sequence members lie inside this epsilonneighbourhood of “a” eventually.
17 00:01:00,340 –> 00:01:03,750 and now we will talk about what happens with the convergence
18 00:01:03,770 –> 00:01:06,510 when we combine two convergent sequences.
19 00:01:07,070 –> 00:01:10,460 and then we also get some rules how to deal with limits.
20 00:01:11,300 –> 00:01:14,480 All of these things we can put into a nice theorem.
21 00:01:15,770 –> 00:01:18,460 Lets simply call it the Theorem on limits.
22 00:01:19,020 –> 00:01:20,060 This simply means
23 00:01:20,080 –> 00:01:23,760 there exists an uniquely given limit “a” for this sequence
24 00:01:23,780 –> 00:01:27,280 and an uniquely given “b” for that sequence.
25 00:01:27,850 –> 00:01:31,380 Ok, now maybe lets visualize these sequences with a graph.
26 00:01:32,140 –> 00:01:35,740 Therefore for example here we can draw the sequence “an”.
27 00:01:35,770 –> 00:01:39,900 So this is a1, a2, a3, a4 and so on.
28 00:01:40,350 –> 00:01:42,840 and in the same way we can look at “bn”
29 00:01:42,860 –> 00:01:45,340 So b1, b2 and so on.
30 00:01:45,870 –> 00:01:48,950 and there you see we can easily form a new sequence
31 00:01:48,970 –> 00:01:52,250 just by combining the two sequences here at each point.
32 00:01:53,050 –> 00:01:55,420 For example we could just add the values
33 00:01:55,450 –> 00:01:57,300 and then we get a new sequence
34 00:01:57,310 –> 00:01:58,610 we could call the sum.
35 00:01:59,190 –> 00:02:01,280 Now of course the natural question is
36 00:02:01,290 –> 00:02:03,390 what happens when “n” goes to infinity?
37 00:02:03,930 –> 00:02:07,920 Because we already know we have convergence for “an” and “bn”
38 00:02:08,389 –> 00:02:13,270 Therefore it would be nice to have it for “an + bn”, the new sequence, as well.
39 00:02:13,980 –> 00:02:16,660 and indeed, this is our first result.
40 00:02:17,210 –> 00:02:22,110 We have that the new sequence “an + bn” is again convergent.
41 00:02:22,280 –> 00:02:24,680 So this limit is uniquely defined.
42 00:02:25,070 –> 00:02:26,550 But we have even more.
43 00:02:26,560 –> 00:02:28,329 We can immediately calculate it.
44 00:02:28,890 –> 00:02:32,270 It’s simply the sum of the two limits we already know.
45 00:02:33,030 –> 00:02:36,790 In other words we can pull in the limit when we have a sum.
46 00:02:37,240 –> 00:02:40,740 Provided that these two limits here actually exist.
47 00:02:41,440 –> 00:02:43,630 Of course this is what we can use a lot
48 00:02:43,640 –> 00:02:47,310 when we want to calculate the limit and we have a plus sign involved.
49 00:02:47,730 –> 00:02:52,350 Now the nice thing is this it works exactly the same when have a multiplication sign
50 00:02:53,060 –> 00:02:56,660 Also here we can just pull the limit into the product.
51 00:02:57,110 –> 00:03:01,020 In other words this new sequence here is also convergent,
52 00:03:01,030 –> 00:03:03,600 when we put in two convergent sequences.
53 00:03:04,400 –> 00:03:09,100 Ok. I can tell you the same thing also holds for the inverse operations
54 00:03:09,160 –> 00:03:11,350 namely minus and division.
55 00:03:11,950 –> 00:03:15,900 So maybe lets write that down for the new sequence given by this fraction.
56 00:03:16,560 –> 00:03:20,500 Of course there we have to put in that the denominator is never 0
57 00:03:20,520 –> 00:03:22,990 such that this whole fraction is well defined.
58 00:03:23,570 –> 00:03:28,079 But then we can just write the limit in the numerator and denominator
59 00:03:28,090 –> 00:03:31,490 provided that this limit here is also not 0.
60 00:03:31,970 –> 00:03:36,560 However under these two assumptions we get the same result as for the product.
61 00:03:37,440 –> 00:03:40,730 Ok, with this we have the 3 important limit theorems.
62 00:03:40,980 –> 00:03:42,170 You really should remember.
63 00:03:42,700 –> 00:03:47,180 We can simply prove them by using our epsilon, “N” definition as always.
64 00:03:47,960 –> 00:03:51,860 Since the proof works in a similar way as we have done it for other proofs before
65 00:03:51,890 –> 00:03:53,280 i can just skip it here.
66 00:03:53,840 –> 00:03:58,040 However it’s really not so hard so we better use the time for a typical example.
67 00:03:58,740 –> 00:04:03,060 So lets take this new sequence “Cn” given by this fraction for all “n”
68 00:04:03,480 –> 00:04:04,750 and ask the question
69 00:04:04,770 –> 00:04:07,150 is this one a convergent sequence?
70 00:04:07,760 –> 00:04:09,850 and if yes, what is the limit?
71 00:04:10,310 –> 00:04:12,740 At first glance this is not clear at all,
72 00:04:12,750 –> 00:04:15,000 because we don’t have the form from above.
73 00:04:15,130 –> 00:04:17,720 We don’t have a convergent sequence here or here.
74 00:04:18,459 –> 00:04:22,740 Indeed if “n” increases this number here gets larger and larger
75 00:04:22,990 –> 00:04:26,250 However what you should see is that the term with “n^2”
76 00:04:26,270 –> 00:04:28,850 is the most important part for this growth.
77 00:04:29,460 –> 00:04:31,820 In other words if “n” is very large
78 00:04:31,840 –> 00:04:35,350 only these two parts here are important for the whole calculation.
79 00:04:35,760 –> 00:04:37,400 Therefore the question remains
80 00:04:37,409 –> 00:04:39,780 how can we deal with this in the correct way.
81 00:04:40,400 –> 00:04:42,400 In fact this always works the same
82 00:04:42,409 –> 00:04:46,320 we just have to expand the whole fraction by this highest power.
83 00:04:46,770 –> 00:04:52,850 Hence in this case just the multiply numerator and denominator by 1 over “n^2”
84 00:04:53,600 –> 00:04:55,800 Then you see everything cancels out
85 00:04:55,830 –> 00:04:58,600 and we get out sequences we already know.
86 00:04:59,070 –> 00:05:00,860 So here constant 2
87 00:05:00,870 –> 00:05:02,800 then 5 over “n” and
88 00:05:02,810 –> 00:05:04,310 1 over “n^2”
89 00:05:04,830 –> 00:05:07,370 and similarly in the denominator.
90 00:05:08,050 –> 00:05:13,150 Now please recall we have already shown that 1 over “n” goes to 0.
91 00:05:13,670 –> 00:05:16,270 So it’s convergent with limit 0.
92 00:05:16,820 –> 00:05:19,690 Then by using the limit theorem part b
93 00:05:19,710 –> 00:05:22,950 we also know that 1 over “n^2” goes to 0.
94 00:05:23,680 –> 00:05:26,340 Therefore by combining all the limit theorems
95 00:05:26,400 –> 00:05:29,890 we see that these parts here all go to 0.
96 00:05:29,980 –> 00:05:34,400 and the only thing that remains in the limit is 2 divided by 5.
97 00:05:35,010 –> 00:05:40,440 More concretely here after the limit we get 2 + 0  0
98 00:05:40,460 –> 00:05:44,050 and here 5 + 0 + 0.
99 00:05:44,760 –> 00:05:48,950 Hence 2 over 5 is indeed the limit of “Cn”
100 00:05:49,800 –> 00:05:54,260 To put it into other words we have proven that this sequence is indeed convergent
101 00:05:54,280 –> 00:05:57,659 without using the definition with epsilon and “N”.
102 00:05:58,640 –> 00:06:02,220 Hence you see the limit theorems make your life much easier here.
103 00:06:02,920 –> 00:06:05,020 Now besides all these properties
104 00:06:05,030 –> 00:06:08,110 the limit also fulfills a monotonicity property
105 00:06:08,490 –> 00:06:12,360 also this is then very helpful for actually calculating limits.
106 00:06:12,840 –> 00:06:17,320 Therefore i show you that in the next video and then we consider more examples.
107 00:06:17,890 –> 00:06:20,470 Therefore i hope i see you there and have a nice day.
108 00:06:20,890 –> 00:06:21,370 Bye!

Quiz Content
Q1: All the things below are valid notations of limits. However, one of them is a false statement. Which one is it?
A1: $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n} = 0$
A2: $\displaystyle \frac{1}{n^2} \xrightarrow{n\rightarrow \infty} 0$
A3: $\displaystyle \frac{1}{n^3} \rightarrow 0$ as $n \rightarrow \infty$
A4: $\displaystyle \frac{1}{n^4}$ tends to $1$ as $n$ tends to infinity
Q2: Let $( a_n ){n \in \mathbb{N}}$ and $( b_n ){n \in \mathbb{N}}$ be convergent sequences. What is not a correct application of the limit theorems?
A1: $\displaystyle \lim_{n \rightarrow \infty} (a_n + b_n) = \lim_{n \rightarrow \infty} a_n + \lim_{n \rightarrow \infty} b_n$
A2: $ \displaystyle \lim_{n \rightarrow \infty} {\Big( (a_n )^3 \Big)} = \Big(\lim\limits_{n \rightarrow \infty} a_n\Big)^3$
A3: $ \displaystyle \lim_{n \rightarrow \infty} f({a_n}) = f\Big(\lim\limits_{n \rightarrow \infty} a_n\Big)$ for a function $f: \mathbb{R} \rightarrow \mathbb{R}$
A4: $\displaystyle \lim_{n \rightarrow \infty} (a_n \cdot b_n) = \lim_{n \rightarrow \infty} a_n \cdot \lim_{n \rightarrow \infty} b_n$
Q3: Which of the following calculations is correct?
A1: $\displaystyle \lim_{n \rightarrow \infty}( (1)^n ) = \pm 1$
A2: $ \displaystyle \lim_{n \rightarrow \infty} (n + (1n) )$ $\displaystyle = \lim_{n \rightarrow \infty} n + \lim_{n \rightarrow \infty} (1n) = \infty  \infty = 0$
A3: $\displaystyle \lim_{n \rightarrow \infty} \frac{ 4 + \frac{1}{n} }{2 + \frac{1}{n} }$ $ \displaystyle = \frac{ \lim_{n \rightarrow \infty} (4 + \frac{1}{n} )}{\lim_{n \rightarrow \infty} (2 + \frac{1}{n} ) } = 2$
A4: $\displaystyle \lim_{n \rightarrow \infty} ( n \cdot \frac{1}{n}) = \lim_{n \rightarrow \infty} n \cdot \lim_{n \rightarrow \infty}\frac{1}{n} = 0$
Q4: Consider the following sequence $(c_n)_{n \in \mathbb{N}}$. What is a suitable step in order to calculate the limit? $$ c_n = \frac{n^4 + 3 n^3 + 2 n^2 1}{6 n^4 + 36}$$
A1: Expand the fraction by $\frac{1}{36}$.
A2: Expand the fraction by $\frac{1}{n}$.
A3: Expand the fraction by $\frac{1}{n^2}$.
A4: Expand the fraction by $\frac{1}{n^3}$.
A5: Expand the fraction by $\frac{1}{n^4}$.