1
00:00:00,000 --> 00:00:06,920

2
00:00:06,920 --> 00:00:07,410
PROFESSOR: Hi.

3
00:00:07,410 --> 00:00:09,150
Welcome back to recitation.

4
00:00:09,150 --> 00:00:11,990
In lecture, you learned about
some new tools for computing

5
00:00:11,990 --> 00:00:12,650
certain limits.

6
00:00:12,650 --> 00:00:15,720
In particular, you learned
about l'Hopital's rule.

7
00:00:15,720 --> 00:00:18,520
So now you have the ability to
compute easily some limits

8
00:00:18,520 --> 00:00:20,410
that before might have
been difficult to do.

9
00:00:20,410 --> 00:00:25,430
So I have here four examples
of limits, some of which--

10
00:00:25,430 --> 00:00:28,150
or maybe all of which, or maybe
none of which or, well

11
00:00:28,150 --> 00:00:31,850
definitely some of which--
will be, you could use

12
00:00:31,850 --> 00:00:33,110
l'Hopital's rule on.

13
00:00:33,110 --> 00:00:34,440
So here they are.

14
00:00:34,440 --> 00:00:38,875
Why do you pause the video and
take a few minutes to work

15
00:00:38,875 --> 00:00:41,000
them all out, come back, and we
can work them out together.

16
00:00:41,000 --> 00:00:49,890

17
00:00:49,890 --> 00:00:50,640
Welcome back.

18
00:00:50,640 --> 00:00:53,420
So we have these four
limits here.

19
00:00:53,420 --> 00:00:56,660
And let's start solving
them, shall we?

20
00:00:56,660 --> 00:01:01,320
So I'll just go from the first
one, and then I'll, you know,

21
00:01:01,320 --> 00:01:02,500
go through them in order.

22
00:01:02,500 --> 00:01:08,205
So limit a is the limit as x
goes to 1 of x to the a minus

23
00:01:08,205 --> 00:01:10,590
1 over x to the b minus 1.

24
00:01:10,590 --> 00:01:13,690
If we want to use l'Hopital's
rule, we have to first check

25
00:01:13,690 --> 00:01:16,300
that this is a limit of
an appropriate kind of

26
00:01:16,300 --> 00:01:16,880
expression.

27
00:01:16,880 --> 00:01:21,870
So that's, in particular, a
limit that is a 0 over 0 or

28
00:01:21,870 --> 00:01:26,500
infinity over infinity form, so
an indeterminate quotient.

29
00:01:26,500 --> 00:01:30,220
And as x goes to 1, we see that
x to the a goes to 1.

30
00:01:30,220 --> 00:01:32,300
So x to the a minus
1 goes to 0.

31
00:01:32,300 --> 00:01:34,950
And similarly x to the b goes
to 1, so x to the b

32
00:01:34,950 --> 00:01:36,360
minus 1 goes to 0.

33
00:01:36,360 --> 00:01:40,780
So indeed, this is a 0 over 0
indeterminate form, so we can

34
00:01:40,780 --> 00:01:42,230
apply l'Hopital's rule.

35
00:01:42,230 --> 00:01:44,590
So we apply l'Hopital's rule.

36
00:01:44,590 --> 00:01:50,950
And it says that the limit as
x goes to 1 of x to the a

37
00:01:50,950 --> 00:01:54,970
minus 1 over x to
the b minus 1.

38
00:01:54,970 --> 00:01:56,700
OK, so what do we do?

39
00:01:56,700 --> 00:01:59,690
What l'Hopital's rule says is we
can take the derivative of

40
00:01:59,690 --> 00:02:02,750
the top and the derivative of
the bottom, and then the limit

41
00:02:02,750 --> 00:02:07,230
of the left side will be equal
to the limit of the result

42
00:02:07,230 --> 00:02:11,150
provided the result actually has
an existing limit, either

43
00:02:11,150 --> 00:02:14,230
a real number or infinity
or minus infinity.

44
00:02:14,230 --> 00:02:17,710
So in this case, this says that
this expression is equal

45
00:02:17,710 --> 00:02:22,030
to the limits-- so it's the same
limit-- as x goes 1 of--

46
00:02:22,030 --> 00:02:28,830
OK, so the derivative of the
top is ax to the a minus 1,

47
00:02:28,830 --> 00:02:33,680
and the derivative of the bottom
is bx to the b minus 1.

48
00:02:33,680 --> 00:02:36,710
So l'Hopital's rule says that
these two limits are equal

49
00:02:36,710 --> 00:02:39,550
provided that the second
limit exists.

50
00:02:39,550 --> 00:02:42,380
So what is the second limit?

51
00:02:42,380 --> 00:02:43,690
Well, we, OK, we can do some
simplifications here.

52
00:02:43,690 --> 00:02:46,250
Here there wasn't, you know,
a lot of obvious

53
00:02:46,250 --> 00:02:47,220
simplifying to be done.

54
00:02:47,220 --> 00:02:49,640
But here there's an obvious
simplification step.

55
00:02:49,640 --> 00:02:55,080
So this is equal to the
limit as x goes to 1.

56
00:02:55,080 --> 00:02:57,720
So a over b is just
a constant.

57
00:02:57,720 --> 00:03:03,380
And now the powers of x, this
is x to the a minus b.

58
00:03:03,380 --> 00:03:07,350
But as x goes to 1, x to the a
minus b goes to 1 no matter

59
00:03:07,350 --> 00:03:08,410
what a and b are.

60
00:03:08,410 --> 00:03:10,980
So at this point, this isn't an
indeterminate form anymore.

61
00:03:10,980 --> 00:03:13,200
This is nice simple limit
we can just plug in.

62
00:03:13,200 --> 00:03:17,520
So this is just equal
to a over b.

63
00:03:17,520 --> 00:03:21,300
OK so that's part a,
straightforward application of

64
00:03:21,300 --> 00:03:22,260
l'Hopital's rule.

65
00:03:22,260 --> 00:03:24,000
Let's go look it part b.

66
00:03:24,000 --> 00:03:27,630
So part b asks us for the
limit as x goes to 0 of

67
00:03:27,630 --> 00:03:29,890
sine 5x over x.

68
00:03:29,890 --> 00:03:34,320
So I'll put that right
here, maybe I'll

69
00:03:34,320 --> 00:03:37,660
draw a little border.

70
00:03:37,660 --> 00:03:47,550
So the limit as x goes to 0 of
sine of 5x divided by x.

71
00:03:47,550 --> 00:03:53,170
So again there's a question of,
you know, is this limit,

72
00:03:53,170 --> 00:03:54,560
do we know what it is already?

73
00:03:54,560 --> 00:03:54,930
and?

74
00:03:54,930 --> 00:03:56,470
The answer is, maybe we
do and maybe we don't,

75
00:03:56,470 --> 00:03:58,210
but suppose we don't.

76
00:03:58,210 --> 00:04:01,500
So in that case, can we apply
l'Hopital's rule?

77
00:04:01,500 --> 00:04:05,520
Well we check, and as x
goes to 0, sine of 5x.

78
00:04:05,520 --> 00:04:09,690
So as x goes to 0, 5x goes to
0, so sine of 5x goes to 0,

79
00:04:09,690 --> 00:04:11,480
and x goes to 0.

80
00:04:11,480 --> 00:04:14,250
So this is another
0 over 0 form.

81
00:04:14,250 --> 00:04:15,980
So we can apply l'Hopital's
rule.

82
00:04:15,980 --> 00:04:21,930
So l'Hopital's rule says that
this is equal to the limit as

83
00:04:21,930 --> 00:04:23,360
x goes to 0 of--

84
00:04:23,360 --> 00:04:25,220
we take the derivative of the
top and divide by the

85
00:04:25,220 --> 00:04:26,640
derivative of the bottom.

86
00:04:26,640 --> 00:04:32,060
So the derivative of the top
is just 5 cosine 5x.

87
00:04:32,060 --> 00:04:35,930
And the derivative of
the bottom is 1.

88
00:04:35,930 --> 00:04:37,850
So l'Hopital's rule says the
limit here is equal to the

89
00:04:37,850 --> 00:04:40,000
limit here provided
this limit exists.

90
00:04:40,000 --> 00:04:44,300
And now as x goes to 0, well,
OK, 5, 1 is constant.

91
00:04:44,300 --> 00:04:47,840
So we're just looking at the
limit of cosine 5x here.

92
00:04:47,840 --> 00:04:51,750
As x goes to 0, that just
approaches 5 times cosine of

93
00:04:51,750 --> 00:04:52,842
0, which is 1.

94
00:04:52,842 --> 00:04:54,970
So this is equal to-- sorry--

95
00:04:54,970 --> 00:04:56,735
the cosine of the 0 part is 1.

96
00:04:56,735 --> 00:05:01,130
So it's 5 times 1, which is 5.

97
00:05:01,130 --> 00:05:04,360
Now I should note that we didn't
actually really need

98
00:05:04,360 --> 00:05:05,810
l'Hopital's rule for this one.

99
00:05:05,810 --> 00:05:09,430
One thing you can remember is
that this limit is actually

100
00:05:09,430 --> 00:05:12,180
the definition of the derivative
of sine of

101
00:05:12,180 --> 00:05:14,390
5x at x equals 0.

102
00:05:14,390 --> 00:05:18,680
So what we just did with
l'Hopital's rule was use the

103
00:05:18,680 --> 00:05:22,040
derivative to compute the
derivative, if you like.

104
00:05:22,040 --> 00:05:24,850
So we used the fact that we
already knew the derivative in

105
00:05:24,850 --> 00:05:26,200
order to compute it.

106
00:05:26,200 --> 00:05:28,790
We could also just say by
definition, this is the

107
00:05:28,790 --> 00:05:34,610
derivative of sine 5x at 0,
which is 5 cosine of 0, which

108
00:05:34,610 --> 00:05:35,820
is 1, which is 5, rather.

109
00:05:35,820 --> 00:05:38,940
OK so that's part b.

110
00:05:38,940 --> 00:05:48,880
Now part c asks for the limit
as x goes to 0 of x squared

111
00:05:48,880 --> 00:05:53,670
minus 6x plus 2-- did
do that right--

112
00:05:53,670 --> 00:05:57,130
over x plus 1.

113
00:05:57,130 --> 00:06:01,090
So this limit, you can't
apply l'Hopital's rule?

114
00:06:01,090 --> 00:06:02,510
Why can't you apply
l'Hopital's rule?

115
00:06:02,510 --> 00:06:05,290
Well, because it's not an
indeterminate form.

116
00:06:05,290 --> 00:06:08,290
This limit is really
easy to compute.

117
00:06:08,290 --> 00:06:10,460
You can just plug in
x equals 0 here.

118
00:06:10,460 --> 00:06:13,560
The top is going to 2, and
the bottom is going to 1.

119
00:06:13,560 --> 00:06:16,365
So the whole limit is going to
2 divided by 1, which is 2.

120
00:06:16,365 --> 00:06:22,080
So there's no, not only, you
don't need l'Hopital's rule,

121
00:06:22,080 --> 00:06:24,350
and also you can't apply
l'Hopital's rule.

122
00:06:24,350 --> 00:06:27,380
This is not a form in which
l'Hopital's rule applies.

123
00:06:27,380 --> 00:06:33,010
It's not an indeterminate form,
OK, so, but OK, but the

124
00:06:33,010 --> 00:06:35,730
result is very easy
to compete.

125
00:06:35,730 --> 00:06:36,740
So that's 2.

126
00:06:36,740 --> 00:06:46,850
And we've got our last one now
which is d, the limit as x

127
00:06:46,850 --> 00:06:55,720
goes to infinity of ln of 1
plus e to the 3x quantity

128
00:06:55,720 --> 00:07:01,690
divided by 2x packs plus 5.

129
00:07:01,690 --> 00:07:05,620
OK, well is it obvious
what this limit is?

130
00:07:05,620 --> 00:07:08,090
Not to me.

131
00:07:08,090 --> 00:07:11,560
So OK, so then we can check.

132
00:07:11,560 --> 00:07:12,810
It's a quotient.

133
00:07:12,810 --> 00:07:14,290
So can we apply l'Hopital's
rule?

134
00:07:14,290 --> 00:07:15,890
Well we can apply l'Hopital's
rule if it's an

135
00:07:15,890 --> 00:07:17,640
indeterminate quotient.

136
00:07:17,640 --> 00:07:20,050
So what's happening is
x goes to infinity?

137
00:07:20,050 --> 00:07:21,780
Well, let's do the bottom first,
because that's easy.

138
00:07:21,780 --> 00:07:25,780
As x goes to infinity, 2x
plus 5 goes to infinity.

139
00:07:25,780 --> 00:07:27,850
And, OK, as x goes
to infinity, e to

140
00:07:27,850 --> 00:07:30,830
the 3x goes to infinity.

141
00:07:30,830 --> 00:07:33,480
So 1 plus e to the 3x
goes to infinity.

142
00:07:33,480 --> 00:07:37,390
So log of 1 plus 3 to the
3x goes to infinity.

143
00:07:37,390 --> 00:07:39,980
So this is an infinity
over infinity form.

144
00:07:39,980 --> 00:07:42,530
That's an indeterminate form,
an indeterminate quotient.

145
00:07:42,530 --> 00:07:44,390
So we can apply l'Hopital's
rule.

146
00:07:44,390 --> 00:07:48,030
So l'Hopital's rule says that
this is equal to the limit of

147
00:07:48,030 --> 00:07:53,920
the quotient of the ratio of the
derivatives, provided that

148
00:07:53,920 --> 00:07:55,590
that limit exists.

149
00:07:55,590 --> 00:08:03,010
So this is equal to the limit
as x goes to infinity of--

150
00:08:03,010 --> 00:08:05,470
OK, so we need to look at the
ratio of the derivatives.

151
00:08:05,470 --> 00:08:07,620
So the bottom one is easy.

152
00:08:07,620 --> 00:08:08,560
That's 2.

153
00:08:08,560 --> 00:08:11,720
For the top one, to compute this
derivative, we need to

154
00:08:11,720 --> 00:08:12,730
use the chain rule here.

155
00:08:12,730 --> 00:08:15,610
So you take the derivative
of this, and you

156
00:08:15,610 --> 00:08:17,270
get, well, a log.

157
00:08:17,270 --> 00:08:23,240
So on the bottom we get
1 plus e to the 3x.

158
00:08:23,240 --> 00:08:30,050
And then by the chain rule, up
top you get 3e to the 3x.

159
00:08:30,050 --> 00:08:32,980
Or, just rewriting this a little
bit in a nicer form,

160
00:08:32,980 --> 00:08:43,480
this is the limit as x goes to
infinity of 3 over 2 times e

161
00:08:43,480 --> 00:08:49,920
to the 3x over 1 plus
e to the 3x.

162
00:08:49,920 --> 00:08:52,190
All right, so now the question
is what do we do from here?

163
00:08:52,190 --> 00:08:55,560
Well is this limit obvious?

164
00:08:55,560 --> 00:08:59,470
Well, OK, so there are
two situations here.

165
00:08:59,470 --> 00:09:02,250
One is, you might look at this
and you might already know

166
00:09:02,250 --> 00:09:03,450
what this limit is.

167
00:09:03,450 --> 00:09:05,390
The reason you might know
that, is that x goes to

168
00:09:05,390 --> 00:09:10,270
infinity, e to the 3x is going
to infinity, and 1 plus e to

169
00:09:10,270 --> 00:09:11,610
the 3x is going to infinity.

170
00:09:11,610 --> 00:09:15,370
So this is an infinity over
infinity indeterminate form.

171
00:09:15,370 --> 00:09:17,275
And so you could apply
l'Hopital's rule to it again.

172
00:09:17,275 --> 00:09:19,280
OK?

173
00:09:19,280 --> 00:09:20,820
That's one thing you could do.

174
00:09:20,820 --> 00:09:23,190
You might also look at and say
well, these two things are

175
00:09:23,190 --> 00:09:25,375
going to infinity at exactly
the same rate.

176
00:09:25,375 --> 00:09:26,380
Right?

177
00:09:26,380 --> 00:09:29,630
This plus 1 is almost
totally irrelevant.

178
00:09:29,630 --> 00:09:32,450
Actually for purposes of
estimating the magnitude of

179
00:09:32,450 --> 00:09:36,170
this fraction, it is
totally irrelevant.

180
00:09:36,170 --> 00:09:38,960
When you take you know, when 3x
is big, then e to the 3x is

181
00:09:38,960 --> 00:09:40,900
like a billion or something.

182
00:09:40,900 --> 00:09:42,320
So e to 3x is a billion.

183
00:09:42,320 --> 00:09:44,760
So you have a billion over
a billion and 1.

184
00:09:44,760 --> 00:09:46,640
So it's very, very close to 1.

185
00:09:46,640 --> 00:09:50,640
It's going to get closer
and closer to 1.

186
00:09:50,640 --> 00:09:53,244
So, OK, so if you didn't trust
that analysis based on

187
00:09:53,244 --> 00:09:57,810
magnitudes, I mean the idea
here is just that the most

188
00:09:57,810 --> 00:10:00,300
significant term in the top
is the e to the 3x.

189
00:10:00,300 --> 00:10:02,810
The most significant term on
the bottom is e to the 3x.

190
00:10:02,810 --> 00:10:05,610
They dominate everything else
that appears, which is just

191
00:10:05,610 --> 00:10:06,590
this little 1.

192
00:10:06,590 --> 00:10:07,990
So that's going to give
you a ratio of 1.

193
00:10:07,990 --> 00:10:10,020
If you don't trust that
analysis, you can apply

194
00:10:10,020 --> 00:10:12,320
l'Hopital's rule again, and
l'Hopital's rule will tell you

195
00:10:12,320 --> 00:10:14,120
exactly the equivalent thing.

196
00:10:14,120 --> 00:10:16,840
When you take a derivative,
you'll be left with 3e to the

197
00:10:16,840 --> 00:10:19,980
3x over 3e to the 3x,
which is just 1.

198
00:10:19,980 --> 00:10:21,980
So in either case, this
part goes to 1.

199
00:10:21,980 --> 00:10:24,430
So three 3/2 times it
goes right to 3/2.

200
00:10:24,430 --> 00:10:27,410

201
00:10:27,410 --> 00:10:29,360
All right, so there
you have it.

202
00:10:29,360 --> 00:10:32,160
Four limits computed with
l'Hopital's rule.

203
00:10:32,160 --> 00:10:35,050
Well one of them was computed
without l'Hopital's rule.

204
00:10:35,050 --> 00:10:37,940
Remember, you always have to
check that you actually have

205
00:10:37,940 --> 00:10:43,400
an indeterminate form before
you apply l'Hopital's rule.

206
00:10:43,400 --> 00:10:44,860
And that's all I have
to say about that.

207
00:10:44,860 --> 00:10:46,560
So I'll end there.

208
00:10:46,560 --> 00:10:46,780