1 00:00:00,012 --> 00:00:07,300 [music] There are plenty of limits that are really quite hard to evaluate. 2 00:00:07,300 --> 00:00:13,278 Here's an example. Take the limit as x approaches 4 of the 3 00:00:13,278 --> 00:00:17,780 square root of x minus 2, divided by x minus 4. 4 00:00:17,780 --> 00:00:22,221 Admittedly, that limit's not really that difficult to calculate. 5 00:00:22,221 --> 00:00:23,716 Alright? The limit is 1 4th. 6 00:00:23,716 --> 00:00:27,571 But we can also think about this, maybe a little bit more intuitively. 7 00:00:27,571 --> 00:00:31,091 Look at what's going on. The limit of the numerator is 0, right? 8 00:00:31,091 --> 00:00:35,336 As x approaches 4, the square root of a number close to 4 minus 2 is close to 0. 9 00:00:35,336 --> 00:00:37,686 And the limit of the denominator is also 0. 10 00:00:37,686 --> 00:00:41,850 And because the limit of the denominator is 0, I can't just use my limit of the 11 00:00:41,850 --> 00:00:45,866 quotients as the quotient of limits deal to evaluate this limit, right? 12 00:00:45,866 --> 00:00:50,966 I'm genuinely in a difficult situation.. And the problem is that the numerator 13 00:00:50,966 --> 00:00:54,862 being close to zero, is trying to make this ratio small. 14 00:00:54,862 --> 00:01:00,772 But when the denominator is close to zero, it's trying to make this ratio really big. 15 00:01:00,772 --> 00:01:04,947 And neither the numerator nor the denominator win the game. 16 00:01:04,947 --> 00:01:08,159 Right? Those 2 forces balance off, giving the 17 00:01:08,159 --> 00:01:09,818 answer of 1 4th. But why? 18 00:01:09,818 --> 00:01:15,712 Well, we can see this geometrically. So here I've graphed in green the function 19 00:01:15,712 --> 00:01:18,826 squared of x minus 2. That's the numerator. 20 00:01:18,826 --> 00:01:22,801 And in orange I've graphed x minus 4. Y equals x minus 4. 21 00:01:22,801 --> 00:01:28,636 And you can see that when x is equal to 4, these curves cross the x axis because when 22 00:01:28,636 --> 00:01:32,908 I plug in 4 here, I get zero. When I plug in 4 here, I get zero. 23 00:01:32,908 --> 00:01:38,290 But also note that the green curve is getting closer to zero more quickly than 24 00:01:38,290 --> 00:01:42,618 the orange curve. Let's replace those two functions with the 25 00:01:42,618 --> 00:01:46,370 approximations that we get by using the derivative. 26 00:01:46,370 --> 00:01:52,250 Now, let me replace this green curve with the tangent line to the green curve at the 27 00:01:52,250 --> 00:01:55,711 point where the green curve crosses the x-axis. 28 00:01:55,711 --> 00:01:59,542 Well, here I go. Here's a picture just of that tangent 29 00:01:59,542 --> 00:02:02,850 line. Now I could compute that tangent line by 30 00:02:02,850 --> 00:02:05,746 doing a little bit of calculus. Alright. 31 00:02:05,746 --> 00:02:08,426 This function is the square root of x minus 2. 32 00:02:08,426 --> 00:02:12,976 If I differentiate that I get 1 over 2 square root sub x, that's the derivative 33 00:02:12,976 --> 00:02:17,051 of the square root function. And the derivative of minus 2 is just 0. 34 00:02:17,051 --> 00:02:21,530 Now I can evaluate that derivative at 4, and I get the derivative is 1 4th, and 35 00:02:21,530 --> 00:02:25,786 that's the slope of the tangent line. Here to the green curve. 36 00:02:25,786 --> 00:02:30,806 Now I know that my tangent line should cross through the point 4, 0. 37 00:02:30,806 --> 00:02:36,752 So here's equation that tangent line in point slope form and I can rewrite this a 38 00:02:36,752 --> 00:02:41,035 little bit more nicely as y equals a quarter of x minus 1. 39 00:02:41,035 --> 00:02:45,996 So what happens to the limit problem when we replace those expressions by their 40 00:02:45,996 --> 00:02:49,060 tangent lines? So instead of thinking about this 41 00:02:49,060 --> 00:02:54,018 geometrically, I can just look back at this original limit problem, and replace 42 00:02:54,018 --> 00:02:58,932 these pieces by their tangent lines. And we calculated the tangent line to the 43 00:02:58,932 --> 00:03:03,957 graph of the square root of x minus 2. To be one quarter times x minus 4, and the 44 00:03:03,957 --> 00:03:08,568 denominator is already a straight line. So, if I replace the numerator and 45 00:03:08,568 --> 00:03:13,032 denominator by the straight line approximations that are good for inputs 46 00:03:13,032 --> 00:03:17,733 near 4, this original limit problem is transformed into this limit problem. 47 00:03:17,733 --> 00:03:20,805 But this limit problem is really easy to do, right? 48 00:03:20,805 --> 00:03:25,700 I've got an x minus 4 in the numerator and an x minus 4 in the denominator. 49 00:03:25,701 --> 00:03:30,125 So this limit is 1 quarter. It's important to emphasize that we really 50 00:03:30,125 --> 00:03:34,590 haven't done anything new here. We could have calculated this limit even 51 00:03:34,590 --> 00:03:39,630 without thinking about derivatives and tangent lines, but this perspective opens 52 00:03:39,630 --> 00:03:42,653 up a new way of approaching some limit problems. 53 00:03:42,653 --> 00:03:47,452 So here's how I can package this up. Into a situation that's maybe useful more 54 00:03:47,452 --> 00:03:50,314 generally. Suppose I got two functions, I got a 55 00:03:50,314 --> 00:03:54,175 function f and a function g. And let's suppose the limit of f as x 56 00:03:54,175 --> 00:03:58,135 approaches a is 0 and the limit of g of x as x approaches a is also 0. 57 00:03:58,135 --> 00:04:02,749 And just for kicks, let's suppose that f and g are also differentiable, twice 58 00:04:02,749 --> 00:04:06,963 differentiable, you know, I want these to be really nice functions. 59 00:04:06,964 --> 00:04:10,838 Now let's suppose I want to try to calculate the limit of f of x over g of x, 60 00:04:10,838 --> 00:04:13,539 as x approaches a. This might be a difficult limit to 61 00:04:13,539 --> 00:04:16,168 calculate. But what we just saw is that we could try 62 00:04:16,168 --> 00:04:20,260 to understand this a little bit better by replacing f and g by their tangent line 63 00:04:20,260 --> 00:04:23,819 approximations. Another way to say that Is I'm just going 64 00:04:23,819 --> 00:04:28,875 to replace f of x and g of x by their approximations that I get from considering 65 00:04:28,875 --> 00:04:33,236 their derivative, right? So, the limit of f of x over g of x should 66 00:04:33,236 --> 00:04:36,951 be close to this because f of x is close to this, right? 67 00:04:36,951 --> 00:04:41,605 It's the value of f at a plus the derivative of f at a times how much the 68 00:04:41,605 --> 00:04:45,901 input changes when I go from a to x. That's approximately f of x. 69 00:04:45,901 --> 00:04:48,851 And this denominator should be about g of x. 70 00:04:48,851 --> 00:04:53,999 It's g of a plus the derivative of g at a times x minus a, that's about g of x. 71 00:04:53,999 --> 00:04:59,309 So if you believe that the numerator and denominator here are approximately f of x 72 00:04:59,309 --> 00:05:03,630 and g of x, you might then believe that these limits are also equal. 73 00:05:03,630 --> 00:05:08,238 But this is a really great situation to be in because f of a say is equal to 0 if f 74 00:05:08,238 --> 00:05:12,012 is continuous there, maybe since it's differentiable. 75 00:05:12,012 --> 00:05:15,242 And g of a is also equal to 0 since g is continuous. 76 00:05:15,242 --> 00:05:18,173 So that these f of a and g of a terms go away. 77 00:05:18,173 --> 00:05:22,911 And then I've got an x minus a and an x minus a term there, so those also cancel, 78 00:05:22,911 --> 00:05:26,691 and all I'm left with is just f prime of a divided by g prime of a. 79 00:05:26,691 --> 00:05:31,113 So it seems like if you try to calculate the limit of this ratio, what's really 80 00:05:31,113 --> 00:05:35,874 relevant is understanding something about the derivatives of your numerator and 81 00:05:35,874 --> 00:05:38,886 denominator separately. At that point a. 82 00:05:38,886 --> 00:05:43,361 The precise statement of this is usually called L'Hopital's rule. 83 00:05:43,361 --> 00:05:48,809 So here's a statement of L'Hopital's rule. Suppose I got two functions f and g and 84 00:05:48,809 --> 00:05:54,303 they're differentiable for inputs near a, and the limit of f of x and the limit of g 85 00:05:54,303 --> 00:05:59,190 of x as x approaches a are both 0. So these functions output is very small 86 00:05:59,190 --> 00:06:03,834 when their input is close to a. And the limit of the derivative of f 87 00:06:03,834 --> 00:06:08,395 divided by the derivative of g as x approaches a, just exists. 88 00:06:08,395 --> 00:06:14,203 And the derivative of g is not 0, when I evaluate that derivative for inputs near 89 00:06:14,203 --> 00:06:17,469 a. If all of these conditions are true, then 90 00:06:17,469 --> 00:06:22,253 I get the fantastic conclusion that the limit of f of x over g of x. 91 00:06:22,254 --> 00:06:26,360 Is the limit of the derivative of f over the derivative of g? 92 00:06:26,360 --> 00:06:31,548 And the hope is that this limit is easier to compute then the original limit. 93 00:06:31,548 --> 00:06:35,648 Let's try to use L'Hopital's rule in a much trickier context. 94 00:06:35,648 --> 00:06:41,158 So here's an example, let's compute the limit as x approaches zero of cosine x 95 00:06:41,158 --> 00:06:45,239 minus one minus x quote over two all over x 2 the 4th power. 96 00:06:45,239 --> 00:06:50,324 Now we can do it using L'Hopital's rule. So the limit of the numerator is zero and 97 00:06:50,324 --> 00:06:55,002 the limit of the denominator is zero. And these are really nice functions, so 98 00:06:55,002 --> 00:06:57,838 they're nice enough for us to apply L'Hopital. 99 00:06:57,838 --> 00:07:02,589 So I should calculate the limit of the derivative of the numerator, divided by 100 00:07:02,589 --> 00:07:07,904 the derivative of the denominator and that might help me compute this original limit. 101 00:07:07,904 --> 00:07:11,299 So here we go. The derivative of the numerator is minus 102 00:07:11,299 --> 00:07:16,401 sign x, that's the derivative cosine. Minus zero, that's the derivative of 1. 103 00:07:16,401 --> 00:07:19,605 Minus x, that's the derivative of x squared over 2. 104 00:07:19,605 --> 00:07:22,809 Divided by, what's the derivative of x to the 4th? 105 00:07:22,809 --> 00:07:26,472 It's 4x cubed. Now here, I'm in a situation where the 106 00:07:26,472 --> 00:07:31,392 limit of the numerator is 0, because the limit of sin x as x approaches 0 is 0 and 107 00:07:31,392 --> 00:07:36,289 the limit of x as x approaches 0 is 0. So, the limit of the numerator is 0 and 108 00:07:36,289 --> 00:07:41,142 the limit of the denominator is also 0. So here's the funny thing. 109 00:07:41,142 --> 00:07:46,332 I could try to apply L'Hopital again to understand this limit. 110 00:07:46,332 --> 00:07:49,486 So here we go. I'll use L'Hopital again. 111 00:07:49,486 --> 00:07:54,237 I'll differentiate the numerator. The derivative of minus sin is minus 112 00:07:54,237 --> 00:07:57,252 cosine. And the derivative of x is 1. 113 00:07:57,252 --> 00:08:02,065 So this is minus cosine of x minus 1. Divided by the derivative of the 114 00:08:02,065 --> 00:08:06,771 denominator which is 4 times 3x squared, or 12x squared. 115 00:08:06,771 --> 00:08:12,109 Now what kind of situation am I in here? Well here, the numerator has limit 0 116 00:08:12,109 --> 00:08:15,634 because cosine's limit is 1 and it's attracting 1. 117 00:08:15,634 --> 00:08:20,899 And the limit of the denominator is also 0, so I'm again in a situation where the 118 00:08:20,899 --> 00:08:24,262 numerator and denominator are heading toward 0. 119 00:08:24,262 --> 00:08:28,883 So to understand this, I could again use L'Hopital to look at how quickly these 120 00:08:28,883 --> 00:08:32,342 terms are dying. So I apply L'Hopital again, and L'Hopital 121 00:08:32,342 --> 00:08:35,354 tells me to look at the derivative of the numerator. 122 00:08:35,354 --> 00:08:38,243 The derivative of minus cosine of x, is sine x. 123 00:08:38,243 --> 00:08:43,645 Minus 1, well that's just 0, divided by the limit of the denominator. 124 00:08:43,645 --> 00:08:47,437 Well, what's the derivative of the denominator? 125 00:08:47,437 --> 00:08:51,275 It's 24x. So now I'm again in a situation where the 126 00:08:51,275 --> 00:08:54,748 numerator and the denominator have limit 0. 127 00:08:54,748 --> 00:09:00,985 So I could again apply L'Hopital, and it would tell me to look at the derivative of 128 00:09:00,985 --> 00:09:05,343 the numerator over the derivative of the denominator. 129 00:09:05,344 --> 00:09:11,162 The derivative of the numerator is cosine x and the derivative of the denominator is 130 00:09:11,162 --> 00:09:14,295 24. But now look, the numerator has limit 1, 131 00:09:14,295 --> 00:09:18,275 and the denominator has limit 24, it's just a constant. 132 00:09:18,275 --> 00:09:23,351 So this is a limit of the quotient and the quotient of limit, so this limit is, Is 1 133 00:09:23,351 --> 00:09:26,236 24th. Now, L'Hopital then tells me that this 134 00:09:26,236 --> 00:09:30,056 limit being equal to 1 24th makes this limit equal to 1 24th. 135 00:09:30,056 --> 00:09:35,002 And because this limit exists, it then tells me that this limit is also equal to 136 00:09:35,002 --> 00:09:38,192 1 24th. And because this limit then exists it then 137 00:09:38,192 --> 00:09:40,901 tells me that this limit is equal to 1 24th. 138 00:09:40,901 --> 00:09:54,876 And then because this limit exists it tells me that the original limit is 1 139 00:09:54,876 --> 00:09:56,022 24th.