AP Calculus AB — Evaluating Limits Algebraically Corrected lesson transcript ## 0:00 — Evaluating Limits Algebraically ## 0:04 — From Graphs and Tables to Equations [0:04] Okay, we're going to start off by talking about a bunch of limit laws. [0:07] So today we're talking about, we're starting to figure out limits, not based on graphs, [0:15] not based on tables or plugging in values, based on just given the function, the equation. ## 0:21 — Limit Laws [0:21] Okay, let me go over some of these laws. [0:23] You don't need to write much notes, but we'll refer back to these many times. [0:32] Just let me give you a rough idea what it is. So the first one, [0:37] the limit of the sum of two functions, [0:41] we can find each limit separately. Limit of f plus the limit of g. So the limit of a sum is the sum of limits. [0:51] The first five, this is only true if the actual limits exist. [0:58] They equal some number. [1:00] Equal some other number. [1:03] If one of the two does not exist, the law does not apply. [1:09] And then the limit of the difference is the difference of limits. [1:16] Okay, any time you have a constant in front of a function, we can bring that outside the [1:23] limit. [1:23] So for example, limit as x goes to 5 or something of 7 times the square root of x plus 3. [1:34] I'm just making some up. [1:35] If we want to we can just bring this guy outside seven times whatever the limit is [1:50] Number four the limit of a product of two different functions is the product of the limits [1:56] For example the limit as x goes to zero [2:01] square root of x times e to the x [2:05] Anytime you have a product that's equal to the limit of the first function times the [2:14] limit whatever the second function is. [2:20] Again that's only true if this equals some number and this equals some number. [2:27] The limit of the division of two functions is the division of the limits. [2:32] This is only true if both of them equal some number and the bottom one does not equal zero. [2:53] This one, if we have some function to any power, for example limit as x goes to... [3:09] the fifth root of 4 to the power x squared. [3:17] Well, that's not the best example. Let me go with cosine of x. [3:23] The fifth root is what power? One fifth power. [3:29] So that's the fifth root or one fifth power of the limit. [3:46] Number seven, the limit of a constant is just equal to constant. [3:51] Why is that? [3:53] Alright, we have some constant function, no matter what x is. [4:03] Whatever a is, as you approach a from the left and the right, it's always equal to whatever [4:10] that constant is, it is whatever that number is. [4:17] Don't worry about 8 for right now. [4:26] Let's skip the next few. [4:29] Let's turn to the back side. ## 4:33 — Direct Substitution Property [4:33] It says if f is a polynomial or rational function, [4:38] and a is in the domain. [4:44] All you have to do is directly substitute in that number [4:47] into the function. [4:49] We talked a little bit about this last Wednesday. [4:54] Why can we do that? [5:01] Yes. [5:03] This is continuous everywhere. [5:06] This is continuous at every number in its domain. [5:11] Everything except what makes the denominator zero. [5:13] So in fact, we're gonna use direct substitution, [5:16] not only for polynomials and rational functions, [5:19] any continuous function. [5:29] Which are all the functions you know. [5:33] polynomials, rational functions, exponential functions, logarithmic functions, [5:38] all trigonometric functions. As long as that a value is in the domain, we just directly substitute it. [5:51] And then we'll go over these last two some other day. ## 5:57 — Trick 1: Direct Substitution [6:05] Alright. [6:07] Let's pull this tricks... [6:13] for... [6:14] and... [6:17] We're gonna have a big list. [6:21] Okay, so leave a lot of room and then maybe start working on the examples on the next [6:25] page. [6:28] Number one trick is direct substitution. [6:40] This is almost always how we start, every limit. [6:44] Ok, so we tried plugging in the value we are approaching. [7:15] What type of function is this? [7:19] Rational. [7:21] We really don't even need to identify as rational for right now but [7:25] we always just try plugging in the number 3 [7:31] I'd do it on the side somewhere [7:33] 3 squared minus 4 over 3 minus 2 [7:38] what is that, 9 minus 4 [7:41] is 5 over 1 is 5. If you end up with a number that is the limit. Okay done easy [7:51] peasy. Okay why does this work? As long as this number is in the domain of a ## 7:57 — 0/0 Is an Indeterminate Form [7:58] continuous function just directly substitute the limit from the left [8:02] always leave the limit from right will equal the function value. Okay so that's the first trick. [8:13] Next example. [8:27] So this is the same function, but now we're approaching it. [8:32] We're approaching as X goes to two. [8:37] Again, on the side, we always try direct substitution. [8:40] 2 squared minus 4 over 2 minus 2. [8:46] And we get 0 divided by 0. [8:50] What is 0 divided by 0? [8:55] That's fine. [8:56] It's not a number. [8:58] OK. [9:00] However, this is very important. [9:04] Any time direct substitution produces 0/0 in the context of a limit, [9:12] this is called an indeterminate form. [9:26] Okay, why is it called an indeterminate form? [9:29] Let me take a guess. [9:33] Because we cannot determine what the value of the limit is. [9:37] The limit might exist, the limit might not exist. [9:40] We cannot determine that yet. [9:42] So any time you get 0/0, it is an indeterminate form. [9:50] And it means we have to do more work. [9:52] Anybody know the trick for this one? ## 9:53 — Trick 2: Factor and Simplify [9:54] We try factoring. [10:09] In fact, every time you get 0 over 0, that almost always means x minus this number is [10:19] both a factor of the numerator and denominator. [10:23] And once you factor it, we should see this common factor. [10:34] We're going to simplify it. [10:38] Oh, before I do that, let me ask you this. [10:54] If you're just given this function, are we allowed to just simplify this function? [11:05] Are they the same function? [11:09] Is this function the same as this function? [11:16] Definitely not. [11:17] They're way different functions. [11:21] Okay, well not way different. [11:27] This function has a hole in it. This function does not have a hole. [11:34] Okay. [11:36] So if you're just given a rational function, you can't just simplify that changes the function. We've got rid of the hole. [11:43] Totally different function. [11:44] However, [11:47] As long as x is not equal to 2, they are exactly equal. [11:54] We agree with that? [11:55] No matter what number you plug in, like we did 3 last time. [12:00] 3 minus 2 over 3 plus 2 over 3 minus 2. [12:04] All these happen to be 1, but this number always simplifies to this number. [12:08] As long as x is not 2. [12:10] So these are the exact same functions, however, are everywhere except at the value of x equals [12:17] 2. So why can we simplify inside limits? [12:23] Okay so we are going to simplify and say that this is the limit as x approaches [12:29] 2 of x plus 2. The reason we could do that inside of a limit is because when [12:37] we're approaching 2 we don't care what happens at 2. All we care about is [12:43] what's happening from the left and from the right. [12:47] Since we're approaching 2, we're never equal to 2. [12:50] So these are always exactly the same number. [12:52] And we can simplify it. [12:54] So inside limits, common factors, we can simplify it. [12:59] And then we go back to our previous trick, which is what? [13:04] Direct substitution. [13:05] So we plug it in. [13:07] And we get 2 plus 2. [13:09] And the answer is 4. [13:17] Okay, so trick number two is factor and simplify. [13:29] Okay, and we often do this trick when you end up with zero over zero. ## 13:34 — Nonzero/0 and Infinite Limits [13:41] What's the first step we always do? [13:46] Direct substitution. [13:48] And again, I do it on the side. [13:51] Get 5 over 3 minus 3. [13:54] And again, direct substitution, doesn't matter if we're going from the left to the right. [13:57] And we get 5 over 0. [14:04] Is this an indeterminate form? [14:08] This is not an indeterminate form. [14:11] OK? [14:12] Any time you have any number other than 0, divided by 0, it's not an indeterminate form. [14:20] Not indeterminate. [14:27] Okay, whenever you get a number over zero, the answer is either going to be infinity [14:34] or negative infinity. [14:36] Okay, here this rational function has a vertical asymptote of three. [14:44] Once we know it's a number over zero, we now just have to determine which one it is positive [14:50] or negative infinity. [14:51] And how do we figure that out? [14:56] How do we know if it's going up or down or just plug [15:01] in a number very close to 3, but from the left, which [15:06] would be what? [15:07] 2.9. [15:08] 2.9 is fine. [15:11] 5 divided by 2.9 minus 3. [15:15] And all we care about is the sign. [15:17] Is it positive or negative? [15:19] 5 divided by negative 0.1 is this positive or negative? [15:26] Definitely negative. [15:32] So what does that tell us? The answer is negative infinity. [15:43] So again, any time you direct substitution, you get a number over zero. [15:50] The answer is going to be either positive or negative. [15:54] Okay, sort of. [15:57] That's when it's going from the left to the right. [16:00] If it was this question, the limit as x goes to 3 of 5 over x minus 3. [16:11] This is not from the left or from the right, it's from both. [16:15] And now we have to check both sides. [16:17] If they both go to negative infinity, the answer is going to be negative infinity. [16:21] So, here we just have to check what is the limit as x goes to 3 from the right. [16:33] Again you substitute and you get 5 over 0. [16:37] So we just have to determine whether it is positive or negative infinity. [16:41] And we just plug in a number to the right of 0. [16:46] 5 over 3.1 minus 3. [16:50] 5 over.1 positive. [16:55] This is positive. [16:58] This limit is equal to positive infinity. [17:04] So what is the limit as x goes to 3? [17:10] Does not exist. [17:12] From the left it went to negative infinity, from the right it went to positive infinity, [17:17] so this does not exist. [17:29] So let's write down this trick. [17:32] Not really a trick, but if you get by direct substitution a number divided by zero, non-zero [17:48] number, the limit will either equal positive infinity or negative infinity, or does not [18:02] exist. You gotta check both the left and the right. [18:11] Zero over zero is an indeterminate form. Any nonzero number [18:14] divided by zero is not indeterminate. The answer is either positive or [18:19] negative infinity. Or from the left and the right they're opposite. [18:23] Does not exist. ## 18:24 — Trick 4: Multiply by a Conjugate [18:25] What is the first thing we always try? [18:29] Direct substitution. [18:31] Square root of 9 minus 3, over 9 minus 9. [18:36] Zero over zero. [18:38] And what do we call that? [18:41] Indeterminate form. [18:43] Indeterminate form. [18:46] So it means we have to do more work. [18:49] Okay. [18:51] Anybody know the trick for this one? [18:55] Multiply by the conjugate. [18:57] Okay. So. [18:57] We are going to multiply by the conjugate of either [19:09] the numerator or the denominator. [19:15] When do we do this trick? [19:16] Typically, you've got to have two different terms. [19:20] OK. [19:20] Both of them have two terms, but especially when [19:24] it involves a square root. [19:26] OK. [19:26] So we're going to multiply the numerator and denominator by the conjugate of one of them. [19:33] In this case, use the conjugate of the numerator: square root of x plus 3. [19:49] the limit [19:51] as x approaches 9 [19:58] of (square root of x minus 3) [20:00] times (square root of x plus 3) [20:02] over (x minus 9) [20:07] times (square root of x plus 3). [20:10] Okay. [20:12] The other thing is, don't distribute the denominator. [20:15] Just leave it the way it is. [20:17] x minus 9 [20:19] times square root of x plus 3. [20:23] We really only distribute the side that, [20:25] or that, uh, the numerator and denominator that [20:28] we've multi- [20:30] uh, who we chose the [20:32] conjugate of. [20:34] And, when you multiply conjugates, [20:36] the two middle terms always add to 0. [20:43] So we get the limit as x approaches 9 [20:47] of (x minus 9) over (x minus 9)(square root of x plus 3). [20:57] and now what? [20:59] if you try plugging in 9 again [21:03] in direct substitution you still get zero over zero. [21:07] However, we go back from our previous trick. [21:10] Most of the time you get zero over zero, [21:12] we have a common factor. [21:15] In this case, we have a common factor of X minus nine. [21:18] We do simplify. [21:30] We try direct substitution now. [21:33] I get one sixth. [21:56] Okay, so, [22:01] Let's call it trick number four. [22:30] So we multiply the numerator and denominator by the conjugate of either the [22:34] numerator or denominator. [22:48] When do we do this? [22:49] That is typically when there is a square root and two terms in either the numerator or denominator. ## 23:00 — Trick 5: Simplify Complex Fractions [23:05] If we try direct substitution here, we get (3 plus 0) to the negative 1 power minus 3 [23:11] to the negative 1 power, all divided by 0. [23:14] We get 0 over 0. [23:20] What are we going to do here? [23:30] What does this mean to the negative one power? [23:35] One over. [23:35] one over so this is [23:46] okay yes wait no no to the negative one power simply means one divided by [23:57] whatever it is just the reciprocal whatever is inside that exponent okay so [24:07] here we have fractions inside of fractions sometimes we call that complex [24:10] fractions not like a complex number just fractions inside fractions so our trick [24:16] is just to make one fraction we don't want fractions inside of fractions so we [24:22] need a common denominator. Multiply one fraction by 3 plus x over 3 plus x, and the other by 3 over 3. [24:48] 3 minus (3 plus x), over 3 times (3 plus x), all divided by x. [25:03] Now we have one fraction divided by another fraction. [25:08] Let me put that there. [25:10] So we take the numerator, [25:19] which, after distributing the negative, becomes 3 minus 3 minus x. [25:24] And that 3 minus 3 is 0. [25:27] So we get negative x over 3 times (3 plus x). [25:33] And we multiply by the reciprocal of the denominator. [25:37] That's what division is. [25:46] And then what happens? [25:50] We have a common factor. [25:52] So it simplifies. [26:11] Now we directly substitute in. [26:30] And when do I stop writing the word limit? [26:36] We stop right when we actually find the limit or take the limit. [26:39] so I did right here now I'm gonna directly substitute it and I got a [26:46] number this one we don't write the word limit anymore as we say we took the [26:52] limit the limit as it approaches is this everywhere else we got to be writing the [26:57] word limit as X approaches whatever so there it is this is negative one ninth so let's [27:06] Let's call this... [27:11] Trick 5. [27:15] Simplify... [27:19] Complex... [27:21] Fractions... [27:22] Which are... [27:25] Fractions within fractions.