# Information about Real Analysis - Part 5

• Title: Sandwich Theorem

• Series: Real Analysis

• YouTube-Title: Real Analysis 5 | Sandwich Theorem

• Bright video: https://youtu.be/Y6rRSip3QN4

• Dark video: https://youtu.be/AZP1fB-9Sok

• Timestamps
• Subtitle in English

﻿1 00:00:00,340 –> 00:00:03,180 Hello and welcome back to real analysis.

2 00:00:03,960 –> 00:00:08,600 and as always i want to thank all the nice people that support this channel on Steady or Paypal.

3 00:00:09,280 –> 00:00:13,560 In todays part 5 we will talk about the Sandwich theorem for limits.

4 00:00:14,190 –> 00:00:19,590 In order to understand this please recall the limit theorems we have for convergent sequences.

5 00:00:19,940 –> 00:00:21,380 For example they tell us

6 00:00:21,400 –> 00:00:25,700 if we look at the sum of the sequence we get a new convergent sequence

7 00:00:25,710 –> 00:00:28,250 and we also can calculate the limit in this way.

8 00:00:28,830 –> 00:00:31,320 and in a similar way we get it for the product.

9 00:00:31,980 –> 00:00:37,350 In particular for the product we can consider the case that “an” is a constant sequence.

10 00:00:37,790 –> 00:00:41,780 This means that we can pull the constant “a” out of the limit.

11 00:00:42,440 –> 00:00:45,350 Of course this property here we will use a lot.

12 00:00:46,210 –> 00:00:49,210 Now i also want to show you two other properties here.

13 00:00:49,880 –> 00:00:53,480 The first one we can call the monotonicity of the limit.

14 00:00:54,030 –> 00:00:59,030 For this lets consider again two convergent sequences “an” and “bn”.

15 00:00:59,620 –> 00:01:04,099 But now at each point “n”, “bn” should be greater or equal than “an”.

16 00:01:04,730 –> 00:01:06,490 Therefore if you draw the graph

17 00:01:06,500 –> 00:01:10,200 the points for “bn” never lie below the points for “an”

18 00:01:10,830 –> 00:01:14,860 Then as always the question is what happens when “n” goes to infinity.

19 00:01:15,510 –> 00:01:19,810 Of course the answer is for the limit we have the same inequality.

20 00:01:20,520 –> 00:01:24,910 However please be careful only the less or equal sign here remains.

21 00:01:25,500 –> 00:01:28,550 So for example if you have a strict inequality here

22 00:01:28,560 –> 00:01:31,150 you still have the less or equal sign for the limit.

23 00:01:31,730 –> 00:01:33,530 Of course you can see this in the graph.

24 00:01:33,540 –> 00:01:36,280 The points could get closer and closer to each other.

25 00:01:36,740 –> 00:01:38,740 And then in the limit they would be the same.

26 00:01:39,650 –> 00:01:43,550 Now the next important property i call the Sandwich theorem.

27 00:01:44,440 –> 00:01:46,880 There we have the same inequality as before,

28 00:01:46,890 –> 00:01:49,789 but now also a new sequence in the middle.

29 00:01:50,460 –> 00:01:52,300 Lets call the sequence “cn”

30 00:01:52,320 –> 00:01:55,310 and then in the graph we would have the points in between.

31 00:01:55,950 –> 00:02:01,220 Now if you also have the information that the limit of “an” and “bn” are the same.

32 00:02:01,270 –> 00:02:04,960 then we see that the point “cn” can’t do so much.

33 00:02:05,430 –> 00:02:10,729 If “n” goes to infinity they need to go to the same limit as “an” and “bn”

34 00:02:11,370 –> 00:02:16,079 and this is the result. The sequence “cn” is also a convergent sequence,

35 00:02:16,100 –> 00:02:17,690 where we already know the limit.

36 00:02:18,370 –> 00:02:21,620 So you see this result is very helpful for calculations,

37 00:02:21,640 –> 00:02:24,790 because you can use sequences you already know

38 00:02:24,800 –> 00:02:28,200 to deduce that a new sequence is also convergent.

39 00:02:28,590 –> 00:02:31,990 Therefore this result is one we really should prove now.

40 00:02:32,460 –> 00:02:36,760 and of course there we already can use the limit theorems from the last video.

41 00:02:37,340 –> 00:02:42,640 For example if you look at the new sequence given by the differerence of “bn” with “an” .

42 00:02:42,900 –> 00:02:48,100 You know by limit theorems this is a convergent sequence and it has the limit 0.

43 00:02:48,770 –> 00:02:55,370 Simply because of the assumption that the limit of “an” we call “a” is the same as the limit of “bn”

44 00:02:56,110 –> 00:02:59,410 and the other assumption is given by the two inequalities here.

45 00:02:59,530 –> 00:03:03,130 So we know all these numbers here are positive or 0.

46 00:03:03,780 –> 00:03:07,680 and the same holds for the new sequence “cn - an”.

47 00:03:08,440 –> 00:03:10,570 So we know we have non-negative numbers,

48 00:03:10,580 –> 00:03:12,480 but we don’t know the limit yet.

49 00:03:12,880 –> 00:03:16,640 So lets call the new sequence simply “dn” and then we know

50 00:03:17,130 –> 00:03:21,720 “dn” lies exactly between 0 and “bn - an”

51 00:03:22,220 –> 00:03:26,310 Now the assumption for our sandwich theorem here looks a little bit simpler here.

52 00:03:27,170 –> 00:03:31,250 and that’s what we can use to prove that “dn” is indeed convergent.

53 00:03:31,800 –> 00:03:35,329 Now using the definition of convergence for given epsilon

54 00:03:35,340 –> 00:03:40,650 we find a “N” such that for all “n” greater or equal than “N”

55 00:03:40,670 –> 00:03:45,950 we have that |bn - an| is less than epsilon

56 00:03:46,510 –> 00:03:49,200 So this is simply the same statement we have above

57 00:03:49,210 –> 00:03:54,420 saying that the sequence “bn - an” is convergent and has the limit 0.

58 00:03:54,800 –> 00:03:58,100 However here we now can use this inequality

59 00:03:58,110 –> 00:04:00,590 and get a statement for “dn” as well.

60 00:04:00,970 –> 00:04:06,060 So “dn” is less than this absolute value and therefore also less than epsilon.

61 00:04:06,640 –> 00:04:12,220 and of course the actual statement we want is to consider |dn - 0|

62 00:04:12,800 –> 00:04:14,830 and when you read the whole sentence now

63 00:04:14,840 –> 00:04:20,829 you see this is exactly the definition that “dn” is a convergent sequence with limit 0.

64 00:04:21,440 –> 00:04:23,930 Therefore this is our conclusion now.

65 00:04:24,530 –> 00:04:29,560 Ok and from this statement we have to go back to our original sequence “cn”.

66 00:04:30,010 –> 00:04:33,850 Which is not so hard. You see we just have to add “an” again.

67 00:04:34,650 –> 00:04:41,530 and of course by the limit theorems we know that the new sequence “dn + an” is also convergent.

68 00:04:42,050 –> 00:04:44,050 Namely with limit “a”.

69 00:04:44,990 –> 00:04:48,090 However this is now simply our sequence “cn”.

70 00:04:48,750 –> 00:04:51,750 and with this we have shown what we wanted to show.

71 00:04:52,490 –> 00:04:54,790 So the sandwich theorem is proven.

72 00:04:55,740 –> 00:04:59,290 I already told you it’s very important so please remember it

73 00:04:59,300 –> 00:05:01,900 and use it in examples where it is possible.

74 00:05:02,480 –> 00:05:07,780 An important thing to note here is that you don’t need the inequalites to hold for all “n”

75 00:05:07,880 –> 00:05:10,370 It’s sufficient that they hold eventually.

76 00:05:10,610 –> 00:05:14,180 This means it can fail for finitely many sequence members.

77 00:05:14,200 –> 00:05:15,860 This does not change the result.

78 00:05:16,310 –> 00:05:19,710 Obviously because this is just a statement about the limit.

79 00:05:20,250 –> 00:05:24,850 and the limit is just not interested in only finitely many sequence members.

80 00:05:25,260 –> 00:05:28,950 Ok, knowing this lets apply the theorem for an example.

81 00:05:29,640 –> 00:05:36,320 Here the sequence “cn” is given by the square root of “n^2 + 1” - “n”.

82 00:05:36,680 –> 00:05:40,530 and the question as always is, is this a convergent sequence?

83 00:05:40,540 –> 00:05:42,740 and in the case it is what is the limit?

84 00:05:43,320 –> 00:05:46,490 The Problem here is that we can’t answer that immediately

85 00:05:46,510 –> 00:05:50,370 simply because we have two parts here, with the subtraction

86 00:05:50,390 –> 00:05:54,270 where both parts go to infinity when “n” goes to infinity.

87 00:05:54,890 –> 00:05:58,620 Therefore it’s not possible to apply the limit theorems immediately.

88 00:05:59,060 –> 00:06:03,960 However the trick is to get rid of the square root and the minus sign at the same time.

89 00:06:04,430 –> 00:06:07,250 and it just works when we expand it to a fraction

90 00:06:07,270 –> 00:06:11,260 by just multiplying with the same term, but now with a plus sign.

91 00:06:11,830 –> 00:06:15,010 Of course then we have a fraction, but we can handle this,

92 00:06:15,020 –> 00:06:17,440 because there is no minus sign anymore.

93 00:06:17,920 –> 00:06:21,120 and in the numerator even the square root cancels out.

94 00:06:21,860 –> 00:06:25,060 There you just have to do the multiplication in the correct way

95 00:06:25,080 –> 00:06:26,960 and we get out two terms.

96 00:06:27,500 –> 00:06:32,200 So we have the square root squared so it vanishes we just have “n^2 + 1”

97 00:06:32,440 –> 00:06:34,830 and then we have minus “n^2”.

98 00:06:35,710 –> 00:06:39,190 The mixed terms vanish, because we have different signs here.

99 00:06:39,800 –> 00:06:44,200 So you see this is very nice, because these two terms here cancel out.

100 00:06:44,840 –> 00:06:48,420 So only 1 over this positive part here remains.

101 00:06:49,170 –> 00:06:54,030 and now we can use that this square root here is always greater than 0.

102 00:06:54,040 –> 00:06:56,730 No matter which natural number “n” we choose.

103 00:06:57,090 –> 00:07:01,810 and therefore this whole fraction is always less than 1 over “n”

104 00:07:02,490 –> 00:07:06,790 and there you see this is exactly what we want for the sandwich theorem.

105 00:07:07,460 –> 00:07:09,750 So that’s the last part we have to write down.

106 00:07:09,880 –> 00:07:11,670 We have the following sandwich

107 00:07:12,220 –> 00:07:15,920 “cn” is always between 0 and 1 over “n”.

108 00:07:16,430 –> 00:07:20,630 and the left hand side and the right hand side both have the same limit.

109 00:07:21,150 –> 00:07:22,930 Namely just 0.

110 00:07:23,400 –> 00:07:26,980 Therefore also “cn” is convergent with limit 0.

111 00:07:27,650 –> 00:07:29,940 and of course this is our result.

112 00:07:30,500 –> 00:07:33,000 So that’s all for our nice example here,

113 00:07:33,020 –> 00:07:35,000 where you have seen a neat little trick

114 00:07:35,020 –> 00:07:37,600 and also the application of the sandwich theorem.

115 00:07:38,340 –> 00:07:41,330 Therefore please try it for yourself for other examples

116 00:07:41,430 –> 00:07:44,020 and also try out the quiz i have in the description.

117 00:07:44,670 –> 00:07:47,670 and with this i hope to see you in the next video. Bye!

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