1 00:00:00,012 --> 00:00:06,501 [music] We already know how to calculate limits that are a very small number 2 00:00:06,501 --> 00:00:10,938 divided by a very small number a zero over zero form. 3 00:00:10,938 --> 00:00:16,572 So we've seen [unknown] Patel's rule for the situation when limit the numerator 4 00:00:16,572 --> 00:00:22,452 limit of the denominator of both zero, but it turns out that[INAUDIBLE] rule is also 5 00:00:22,452 --> 00:00:27,912 valid when the limit of the numerator and the limit of the denominator are both 6 00:00:27,912 --> 00:00:30,705 infinity. I mean you still got all the other 7 00:00:30,705 --> 00:00:33,526 conditions. You gotta check that the limit of the 8 00:00:33,526 --> 00:00:38,184 ratio of the derivative exists, the derivative of the denominator is not zero. 9 00:00:38,185 --> 00:00:42,725 For values of x near a. But given these conditions, you then get 10 00:00:42,725 --> 00:00:47,765 the same fantastic conclusion. That the limit of f over g is the limit of 11 00:00:47,765 --> 00:00:50,994 the derivative of f over the derivative of g. 12 00:00:50,995 --> 00:00:53,847 Let's take a look at an example. Well, here's an example. 13 00:00:53,847 --> 00:00:57,688 This is an example, you really don't need to use[UNKNOWN] a lot, but at least it 14 00:00:57,688 --> 00:01:01,480 demonstrates what the technique is. The limit on the numerator is infinity, 15 00:01:01,480 --> 00:01:05,308 the limit ofthe denominator is also infinity, though that's maybe a little bit 16 00:01:05,308 --> 00:01:09,039 harder to see. And consequently We could use lopatol/g, 17 00:01:09,039 --> 00:01:13,958 and it would tell us to compute instead the derivative the numerator. 18 00:01:13,958 --> 00:01:19,515 Which is 4 x plus 0, divided by the derivative the denominator, which is 6 x 19 00:01:19,515 --> 00:01:22,447 minus 1. Now can I calculate the limit of this as x 20 00:01:22,447 --> 00:01:25,515 approaches infinity. Yes, of course I could just do this 21 00:01:25,515 --> 00:01:29,102 directly too but I could again use lowbatal of I really wanted to. 22 00:01:29,102 --> 00:01:33,312 The limit of the numerator is infinity, the limit of denominator is infinity. 23 00:01:33,312 --> 00:01:37,734 So lobatall tells me to instead consider the ratio of derivatives, the derivative 24 00:01:37,734 --> 00:01:41,214 of the numerator is 4, the derivative of the denominator is 6. 25 00:01:41,214 --> 00:01:45,311 And now I'm in a situation where I'm just taking the limit of a constant over a 26 00:01:45,311 --> 00:01:48,063 constant. That means that limit is 4 6th, which then 27 00:01:48,063 --> 00:01:52,095 tells me how to compute the original limit, but, and this isn't the best way to 28 00:01:52,095 --> 00:01:56,001 do this problem, but at least it demonstrates that you could use Lopital in 29 00:01:56,001 --> 00:01:58,812 a situation where you got infinity over infinity. 30 00:01:58,813 --> 00:02:01,745 Infinity. But not every limit is something going to 31 00:02:01,745 --> 00:02:06,233 zero over something going to zero, or something going to infinity over something 32 00:02:06,233 --> 00:02:09,461 going to infinity. Sometimes you might have a limit where 33 00:02:09,461 --> 00:02:13,619 it's a product, the first term is heading towards zero and the second term is 34 00:02:13,619 --> 00:02:17,299 heading towards infinity. For example, here I'm asking what the 35 00:02:17,299 --> 00:02:20,655 limit as x approaches infinity of sin of one over x times x. 36 00:02:20,656 --> 00:02:25,033 This first term, sine of 1 / x, that's getting close to 0 when x is very large. 37 00:02:25,033 --> 00:02:28,275 And the second term, just the x, that is getting very big. 38 00:02:28,275 --> 00:02:32,835 So this first term is having a tendency to made this quantity smaller but the second 39 00:02:32,835 --> 00:02:35,950 term is having a tendency to make the quantity bigger. 40 00:02:35,950 --> 00:02:40,227 You know, who wins?. Well with lobitol/g the only thing that we 41 00:02:40,227 --> 00:02:45,427 can really deal with is sort of a 0 over 0 situation or an infinity over infinity 42 00:02:45,427 --> 00:02:51,002 situation and that's neither of these. But we can transform this problem into one 43 00:02:51,002 --> 00:02:55,242 of these situations. So instead of calculating this limit I'll 44 00:02:55,242 --> 00:03:00,113 calculate the equivalent problem, the limit as x approaches infinity. 45 00:03:00,114 --> 00:03:03,362 Of sine 1 over x in the numerator divided by 1 over x. 46 00:03:03,362 --> 00:03:08,222 So instead of multiplying by something which is going towards infinity, I'm going 47 00:03:08,222 --> 00:03:12,325 to divide by its reciprocal. The reciprocal of something going to 48 00:03:12,325 --> 00:03:16,044 infinity, is going to 0. So now I've got a situation where the 49 00:03:16,044 --> 00:03:20,114 numerator is going to 0, and the denominator is going to 0, and this 50 00:03:20,114 --> 00:03:23,103 problem, equivalent to the original problem. 51 00:03:23,103 --> 00:03:28,233 Is now amenable to L'Hospital's rule. So by L'Hospital's rule I would want to 52 00:03:28,233 --> 00:03:33,763 differentiate the numerator, differentiate the denominator and then look at that 53 00:03:33,763 --> 00:03:36,932 limit to try to understand this original limit. 54 00:03:36,932 --> 00:03:42,356 Well, what's the derivative of, Of sine of 1 over x, it's cosine of 1 over x, because 55 00:03:42,356 --> 00:03:47,946 the derivative of sine is cosine, times the derivative of the inside, which is 56 00:03:47,946 --> 00:03:51,686 minus 1 over x squared. And I'm going to divide by the 57 00:03:51,686 --> 00:03:56,423 derivitative of the denominator, which is minus 1 over x squared. 58 00:03:56,424 --> 00:04:01,432 So, now how do I evaluate this limit? Well, the good news is that I've got a 59 00:04:01,432 --> 00:04:06,270 minus 1 over x squared in the numerator and a minus 1 over x squared in the 60 00:04:06,270 --> 00:04:09,896 denominator. So this limit is the same as the limit as 61 00:04:09,896 --> 00:04:13,191 x approaches infinity of just cosine of 1 over x. 62 00:04:13,191 --> 00:04:18,592 But now, 1 over x is getting very close to 0 and what's cosine of a number close to 63 00:04:18,592 --> 00:04:19,413 0? It's 1. 64 00:04:19,413 --> 00:04:24,133 So this original limit, is 1. You might have something near 1, being 65 00:04:24,133 --> 00:04:27,335 raised to a very high power. Here's an example. 66 00:04:27,335 --> 00:04:32,635 The limit as x approaches infinity, of 1 plus 1 over x, raised to the xth power. 67 00:04:32,635 --> 00:04:37,870 So for very large values of x, the base here, 1 plus 1 over x, that's close to 1, 68 00:04:37,870 --> 00:04:42,722 but the exponent is very large. If you take a number close to 1 and raise 69 00:04:42,722 --> 00:04:45,968 it to a very. Very high power it's actually unclear what 70 00:04:45,968 --> 00:04:49,288 you're going to get. Depending as to how quickly this is moving 71 00:04:49,288 --> 00:04:51,997 towards 1 and how quickly this thing is growing. 72 00:04:51,997 --> 00:04:56,098 You get wildly different answers. Now this is not 0 over 0 or infinity over 73 00:04:56,098 --> 00:04:59,033 infinity. So I've got to transform this problem into 74 00:04:59,033 --> 00:05:02,322 something. Thing that L'Hopital can handle. 75 00:05:02,322 --> 00:05:06,027 The trick here is to use exponential functions. 76 00:05:06,027 --> 00:05:11,955 So I'm going to rewrite this as e to the log of the limit as x approaches infinity 77 00:05:11,955 --> 00:05:17,539 of 1 plus 1 over x to the xth power. E to the log of something does nothing, 78 00:05:17,539 --> 00:05:20,525 ritght? These are inverse functions. 79 00:05:20,525 --> 00:05:23,808 But, log of a limit is the limit of the log. 80 00:05:23,808 --> 00:05:28,852 So this is. E to the limit as x approaches infinity of 81 00:05:28,852 --> 00:05:32,820 the log of 1 plus 1 over x to the xth power. 82 00:05:32,820 --> 00:05:39,823 Now log as something to a power is that power times the log, so this is e. 83 00:05:39,824 --> 00:05:45,182 To the limit as X approaches infinity of X times the log of one plus one over X. 84 00:05:45,182 --> 00:05:50,622 Now what kind of situation am I in here? X is very large but log of a number close 85 00:05:50,622 --> 00:05:54,993 to one is close to zero. This is big number times number close to 86 00:05:54,993 --> 00:05:57,562 zero. That's the infinity times zero 87 00:05:57,562 --> 00:06:01,239 indeterminate form, so how am I going to handle this? 88 00:06:01,239 --> 00:06:05,263 We just saw a minute ago that I'm going to handle this by. 89 00:06:05,264 --> 00:06:10,762 Putting the infinity in the denominator with the reciprocals to make this 0 over 90 00:06:10,762 --> 00:06:14,121 0, the sort of thing that[INAUDIBLE] can handle. 91 00:06:14,121 --> 00:06:19,439 So this is e to the limit as x approaches infinity of lg 1 plus 1 over x divided by 92 00:06:19,439 --> 00:06:22,692 1 over x. This is the same as this, but now I've got 93 00:06:22,692 --> 00:06:26,921 something approaching 0 divided by something approaching 0. 94 00:06:26,921 --> 00:06:31,248 This is the sort of situation that[INAUDIBLE] can help me with. 95 00:06:31,249 --> 00:06:36,540 This is e to the limit, by lopital. The derivative of the numerator divided by 96 00:06:36,540 --> 00:06:41,461 the derivative of the denominator. The derivative of log is one over the 97 00:06:41,461 --> 00:06:46,791 inside function, times the derivative of the inside, which is minus one over x 98 00:06:46,791 --> 00:06:50,367 squared. Divided by what's the derivative of 1 over 99 00:06:50,367 --> 00:06:55,295 x, well it's the same thing here minus 1 over x squared and this is the limit as x 100 00:06:55,295 --> 00:06:59,058 approaches infinity. Now I've got a minus 1 over x squared 101 00:06:59,058 --> 00:07:03,262 which cancels with the minus 1 over x squared in the denominator. 102 00:07:03,262 --> 00:07:07,130 And I've got 1 over 1 plus 1 over x as x approaches infinity. 103 00:07:07,131 --> 00:07:12,781 This is getting very close to one. So this is e to the 1st power, which is e. 104 00:07:12,781 --> 00:07:18,338 And that means that this original limit, the limit of 1 plus 1 over x to the x 105 00:07:18,338 --> 00:07:21,847 power as x approaches infinity, is equal to e. 106 00:07:21,847 --> 00:07:27,364 Or you might have a very large number, being raised to a very small power. 107 00:07:27,364 --> 00:07:32,782 For example, let's say I want to come up with the limit as x approaches infinity of 108 00:07:32,782 --> 00:07:37,109 x to the 1 over xth power. So the base here is very large, which have 109 00:07:37,109 --> 00:07:42,244 a tendency to make this number very big, but the exponent is getting close to 0, 110 00:07:42,244 --> 00:07:46,466 which would have a tendency to pull this back down closer to 1. 111 00:07:46,467 --> 00:07:50,671 So what is this? Well we can try to transform this into the 112 00:07:50,671 --> 00:07:56,152 sort of limit problem that[UNKNOWN] can handle, and I can again do that with 113 00:07:56,152 --> 00:08:00,819 exponential functions. So I can rewrite this as e to the log of 114 00:08:00,819 --> 00:08:06,222 the limit of x to the 1 of xth power. And this is the limit as x approaches 115 00:08:06,222 --> 00:08:09,460 infinity. Now, this is the log of a limit. 116 00:08:09,460 --> 00:08:13,730 Which is the limit of the log of x to the 1 over xth power. 117 00:08:13,730 --> 00:08:17,334 And this is the limit as x approaches infinity. 118 00:08:17,334 --> 00:08:21,074 But the log of something to a power is that power. 119 00:08:21,074 --> 00:08:24,812 So 1 over x times. The log of the base as x approaches 120 00:08:24,812 --> 00:08:28,266 infinity. Now I've got 1 over x, which is the number 121 00:08:28,266 --> 00:08:31,643 close to zero. Times log of x, which is a very large 122 00:08:31,643 --> 00:08:34,989 number. This is zero times infinity, so to speak. 123 00:08:34,989 --> 00:08:39,753 So I should try to transform this indeterminate form into something that 124 00:08:39,753 --> 00:08:44,401 labetalol can handle. Well, we'll write this a e to the limit of 125 00:08:44,401 --> 00:08:47,702 x approaching infinity, of say log x over x. 126 00:08:47,702 --> 00:08:51,023 This is infinity over infinity, so to speak. 127 00:08:51,023 --> 00:08:56,749 That's the sort of thing that L'Hopital's okay with, so instead of taking this 128 00:08:56,749 --> 00:08:59,842 limit. I could look at the ratio of the 129 00:08:59,842 --> 00:09:03,786 derivatives. The derivative of log x is 1 over x. 130 00:09:03,786 --> 00:09:08,774 The derivative of x is 1. So I should look at the limit of 1 over x 131 00:09:08,774 --> 00:09:14,506 over 1 as x approaches infinity. Well, that limit is 0 and e to the 0 is 1, 132 00:09:14,506 --> 00:09:20,111 so the limit of x to the 1 over x as x approaches infinity is equal to 1. 133 00:09:20,111 --> 00:09:24,482 Or you might have a limit that looks like something going to infinity minus 134 00:09:24,482 --> 00:09:28,402 something going to infinity. So let's try to compute the limit as x 135 00:09:28,402 --> 00:09:32,366 approaches infinity of the square root of x squared plus x minus x. 136 00:09:32,366 --> 00:09:35,266 So this is a very large number minus a very large. 137 00:09:35,266 --> 00:09:40,046 The infinity minus infinity situation. So we should try to factor this or rewrite 138 00:09:40,046 --> 00:09:43,726 this to get it into a zero over zero or infinity over infinity. 139 00:09:43,726 --> 00:09:46,438 The sort of thing I could apply lopitol/g to. 140 00:09:46,439 --> 00:09:49,802 2. So I could try to pull out an x from this, 141 00:09:49,802 --> 00:09:55,548 because x is going to infinity. I know x is a large positive number here. 142 00:09:55,548 --> 00:10:01,032 So I could rewrite this as x times, so what if I pull out an x from here? 143 00:10:01,032 --> 00:10:06,922 That's the square root of 1 plus 1 over x. Minus, when I pull out an extra one here I 144 00:10:06,922 --> 00:10:10,518 get minus 1. So really, this limit, for large values of 145 00:10:10,518 --> 00:10:15,653 x, so as x approaches infinity, is the same as x times this quantity, the square 146 00:10:15,653 --> 00:10:19,168 root of 1 plus 1 over x minus 1. This is a large number. 147 00:10:19,168 --> 00:10:23,892 What do I know about this number? Well this is the square root of 1 plus the 148 00:10:23,892 --> 00:10:27,600 reciprocal of a large number. This is close to 1 minus 1. 149 00:10:27,600 --> 00:10:33,813 This second term is close to zero. This is infinity times 0 in determinate 150 00:10:33,813 --> 00:10:38,598 form. So I could rewrite this using our standard 151 00:10:38,598 --> 00:10:43,795 trick as 1 plus 1 over x minus 1. So this is now the numerator. 152 00:10:43,795 --> 00:10:46,905 This thing's going to 0 divided by 1 over x. 153 00:10:46,905 --> 00:10:51,353 The reciprocal of this. But I'm dividing by it, and that's the 154 00:10:51,353 --> 00:10:55,531 same as multiplying. So now I've got 0 divided by a number 155 00:10:55,531 --> 00:10:59,027 close to 0. I, it's 0 over 0 in determinate form. 156 00:10:59,027 --> 00:11:03,847 So lopital tells me I can analyze this by looking at the ratio of the, The 157 00:11:03,847 --> 00:11:09,491 derivatives, so I should look at the limit as x approaches infinity of, what's the 158 00:11:09,491 --> 00:11:13,834 derivative of this? Well, the derivative of the square root is 159 00:11:13,834 --> 00:11:18,954 one over two square root of the inside function, one plus one over x, times the 160 00:11:18,954 --> 00:11:23,192 derivative of the inside function. So the derivative of 1 over x. 161 00:11:23,192 --> 00:11:27,546 And I don't have to worry about the minus 1, because the derivitive of that's 0. 162 00:11:27,546 --> 00:11:32,242 And I divide by the derivitave of the denominator, so I'll just write derivative 163 00:11:32,242 --> 00:11:35,752 1 over x. So now I've got derivative 1 over x in the 164 00:11:35,752 --> 00:11:39,660 numerator, derivative 1 over x in the denominator. 165 00:11:39,660 --> 00:11:43,958 This limit then, should be same as just the limit of this. 166 00:11:43,958 --> 00:11:49,634 So I should be looking at the limit as x approaches infinity of 1 over 2 square 167 00:11:49,634 --> 00:11:51,677 root. Of one plus one over x. 168 00:11:51,677 --> 00:11:56,494 Well, what's one plus on over x as x approaches infinity, that's just one. 169 00:11:56,494 --> 00:12:01,458 So this is one over two square root of a number close to one, this is one half. 170 00:12:01,458 --> 00:12:04,926 And, indeed, this original limit really is one half. 171 00:12:04,926 --> 00:12:10,002 I mean, honestly, we didn't need l'hopital to calculate that, but we could use 172 00:12:10,002 --> 00:12:13,131 l'hopital if we wanted to. To evaluate this limit. 173 00:12:13,131 --> 00:12:15,669 Okay, okay. Let's summarize all the possibilities. 174 00:12:15,669 --> 00:12:19,062 So here's this summary of everything you might see in the[INAUDIBLE]. 175 00:12:19,062 --> 00:12:22,530 If you see zero over zero, or infinity over infinity in a limit problem. 176 00:12:22,530 --> 00:12:24,652 You can just apply L'Hopital, in that case. 177 00:12:24,652 --> 00:12:28,036 But here, I've lifted off some of the other things that you might see. 178 00:12:28,036 --> 00:12:31,430 And these in equations, there's no nonsense equations, right? 179 00:12:31,430 --> 00:12:34,133 But I hope they kind of tell you what you should do. 180 00:12:34,133 --> 00:12:38,299 So if you see, say 0 times infinity in a limit, and by that I mean, it's a product 181 00:12:38,299 --> 00:12:42,325 of things, one of which is limit 0, the other of which has limit infinity. 182 00:12:42,325 --> 00:12:46,245 Well, you should transform this into one of these cases so you can apply L'Hopital. 183 00:12:46,246 --> 00:12:51,605 So you could do that by moving the 0 into the denominator, and taking its reciprocal 184 00:12:51,605 --> 00:12:56,408 This is really infinity over infinity. Or, you could move the infinity in the 185 00:12:56,408 --> 00:13:00,336 denominator, and take its reciprocal. And now you've got 0 over 0. 186 00:13:00,336 --> 00:13:04,662 If you see 1 to a very large power, or something close to 1 to a large power, 187 00:13:04,662 --> 00:13:07,356 right? You can use either the log to transform 188 00:13:07,356 --> 00:13:09,966 this. And e to the log of this is e to the large 189 00:13:09,966 --> 00:13:14,247 number times log of number close to 1. Log of a number close to 1 is close to 190 00:13:14,247 --> 00:13:16,770 zero. This is an infinity times 0 indeterminate 191 00:13:16,770 --> 00:13:18,838 form. But you know how to handle those. 192 00:13:18,838 --> 00:13:21,539 Because you can convert them back into these cases. 193 00:13:21,539 --> 00:13:24,935 Which you can then use[INAUDIBLE]. If you see infinity to the 0. 194 00:13:24,935 --> 00:13:27,426 Well, you could use the same e to the log trick. 195 00:13:27,426 --> 00:13:30,398 And then the exponent here is 0 times log of a big number. 196 00:13:30,398 --> 00:13:32,889 Or a number close to 0 times log of a big number. 197 00:13:32,889 --> 00:13:36,741 But this is the 0 times infinity indeterminate form, which is right here. 198 00:13:36,741 --> 00:13:40,650 We should transform you to these cases, which you then use L'Hopital on. 199 00:13:40,651 --> 00:13:45,407 If you see 0 to the 0, by which I mean, a number close to 0 raised to a power close 200 00:13:45,407 --> 00:13:49,982 to 0, you can again use the e to the log trick And this is e to a number close to 201 00:13:49,982 --> 00:13:54,185 0, times log and a number close to 0. What's log of a very small number? 202 00:13:54,185 --> 00:13:57,826 That's very negative. So, this exponent is, again, the 0 times 203 00:13:57,826 --> 00:14:02,434 infinity in determinate form, which you can then convert into this and apply 204 00:14:02,434 --> 00:14:05,887 L'Hopital To it. The last case is this infinity minus 205 00:14:05,887 --> 00:14:11,167 infinity case, and there, one thing you could try to do is put it over some common 206 00:14:11,167 --> 00:14:16,447 denominator, so I'm thinking of the common denominator here as being 1 over the 207 00:14:16,447 --> 00:14:21,647 product of these two terms, and here I've got 1 over the second term minus 1 over 208 00:14:21,647 --> 00:14:25,542 the first term. But the point here is that you can rewrite 209 00:14:25,542 --> 00:14:32,654 this difference of very large quantities. As a something getting close to 0 divided 210 00:14:32,654 --> 00:14:39,537 by something getting close to 0. Which is the sort of thing that you could 211 00:14:39,537 --> 00:14:41,704 then apply L'Hopital.