WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about calculating limits mathematically.
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Let us jump right on in.
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We have been using graphs and we are going to be using tables of values.
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Now we are going to do it analytically.
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Let us go to blue here.
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We have been using graphs and/or tables to find limits.
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How do we do it analytically?
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How do we do it analytically, mathematically?
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If there are some technique that we can use to evaluate this limit.
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For example, what if we have something like the limit as x approaches 2 of x³ + x² - 4/ 6 - 4x.
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What if we are faced with some limit like that?
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How do we deal with this?
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The short answer is the following.
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The short answer is plug 2 into f(x) and see what you get.
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I know, that is it, and see what you get.
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Yes, that is literally it.
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Just plug in and see what you get.
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When we put 2 in here, we end up with 2³ + 2² - 4/ 6 - 4 × 2.
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8 + 4 – 4/ 6 - 8 = 8/-2 = -4.
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That is actually your answer.
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When you plug the value that x approaches into f(x), one of two things happens.
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If you get an actual number like we did, you can stop.
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This is your limit.
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This is the limit.
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If you do not, if you get something that does not make sense, for our purposes,
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does not make sense is going to be something like dividing by 0 or 0/0, infinity/infinity, infinity – infinity,
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things that later on we are going to call indeterminate forms.
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If you get something that does not make sense, you are going to have to manipulate the expression.
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Turn it into something else, take the limit again.
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If you get something that does not make sense, you must try other things.
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Usually what that means, usually other things means, this means rewriting f(x) through some sort of mathematical manipulation.
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You are going to turn it into an equivalent expression, but you are just going to manipulate it mathematically.
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Maybe you are going to simplify something algebraically.
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Maybe you are going to factor and cancel something out.
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You are going to try different things.
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Maybe you are going to rationalize the denominator, rationalize the numerator, whatever it is that you are going to do.
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You are going to convert it to an equal of expression and then take the limit again, until you get something.
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Usually this means rewriting f(x) through some sort of mathematical manipulation.
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Let us see why this plug in technique works.
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I can leave it and just jump right in, but I think it is important to at least see why it works.
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Let us go to blue.
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Let us see why this plug-in procedure works.
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We are going to begin with two very basic limits, very obvious limits.
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We begin with two basic limits that are obvious.
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The limit as x approaches a of a constant c = c.
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The graph of it looks like this.
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This is your coordinate system.
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Let us just say this is f(x) equal c, just some constant.
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Let us say for example, f(x) = 5.
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It does not matter what the value of x is.
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f(x) is always going to equal 5.
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Clearly, no matter what number you approach, the limit of the constant, the limit as x approaches a of 5 is going to be 5.
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This is an obvious limit.
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The limit as x approaches a of a constant is the constant.
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That is our first obvious limit.
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The second obvious limit, let me go back to blue, is the limit as x approaches a of x actually equal a.
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This graph looks like this.
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The graph of the function x looks this way.
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Let us say this is a, as x approaches a, this side and this side.
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This function right here is y = x.
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Therefore, the y value here is also a.
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The y value is approaching a, from above the y value is approaching a.
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This is an obvious limit.
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The limit of the function x as x approaches a = a, = the number you are approaching.
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Once again, it is clear that the limit as x approaches some number a of 5 is equal to 5, that one is clear.
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Hopefully also, this one is also clear that the limit as x approaches some number of the function of x = that number.
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Now that we have those two basic limits, let us write down our limit laws.
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Here we will let c be any constant and the limit as x approaches a of f(x) is a.
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The limit of a function as x approaches a exists and it equals a.
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We will let the limit as x approaches a also of a function g(x), we will let it equal b.
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Our first limit law is the following.
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The limit as x approaches a of f(x) + g(x) = the limit as x approaches a of f(x) + the limit as x approaches a of g(x) = a + b.
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What this says is that, the limit of the sum of two functions is equal to the sum of the individual limits of the functions.
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You can think of the limit as distributing over both.
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The limit of something + something is the limit of something + the limit of the second something.
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We have two functions, you add them, the limit of the sum is the sum of the limits.
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Let us try this one.
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The limit as x approaches a of f – g.
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That is the same thing, this is the same as that except this is just –g.
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It is going to be the limit as x approaches a(f) - the limit as x approaches a(g) which is equal to a – b.
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Because the limit of f(x) is a, the limit of f is a, the limit of g is b.
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The third, the limit as x approaches a of any constant × f(x).
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It is equal to the constant × the limit as x approaches a of f(x), it is equal to c × a.
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The limit of a constant × that, you can pull the constant out and put in front of the limit.
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Number 4, the limit as x approaches a of f(x) × g(x).
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It is exactly what you think, the limit of the product is equal to the product of the individual limits.
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= the limit as x approaches a of f(x) × the limit as x approaches a of g(x) which is equal to a × b.
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We have 5, the limit as x approaches a of f(x)/ g(x) = the limit as x approaches a of f(x)
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divided by the limit as x approaches a of g(x).
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That is it, nice and straightforward.
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The limit of the sum is the sum of the limits.
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The limit of the constant × the function is the constant × the limit.
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The limit of the product is the product of the limits.
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The limit of the quotient is the quotient of the limits.
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Provided it exists down here, and we said that it does.
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Let us do a 4’, the limit as x approaches a of f(x) raised to the n power is equal to the limit as x approaches a of f(x).
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The limit of the function f(x) raised to the n is equal to limit of f(x) all raised to the n.
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It is just f(x) multiplied a certain number of times, that is all it is.
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That equals the limit raised to the n.
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Now applying 4’ to the function f(x) = x.
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We get the limit as x approaches a(x) ⁺n is equal to the limit as x approaches a(x) ⁺n.
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We said that the limit is x approaches a(x) is equal to a.
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It is just equal to a ⁺n.
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That is it, very nice.
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The last limit, the limit as x approaches a of n √f(x) = the n √of the limit as x approaches a of f(x).
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You just pass the limit through.
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Our limit procedure, our plug-in procedure is summarized as,
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given a polynomial function f(x),
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if a is in the domain of f, then the limit as x approaches a of f(x) = f(a).
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I just wrote this down for the sake of writing it down.
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From a practical standpoint, I would not worry about this.
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The truth is whether a is in the domain or not, we are just going to plug in a.
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We are going to see what we get.
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If it is in the domain, it will work out just fine, we will get a number.
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If something else happens, like we said before, you are going to have to manipulate it to see what you can do with it.
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I would not worry about what it is that I just wrote.
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The idea is basically just plug in and see what you get.
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Let us go back to that limit that we started off with.
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Sorry, was it 2 or was it 5?
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I think it was 2.
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The limit as x approaches 2 of, we had x³ + x² - 4/ 6 - 4x.
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The shortcut was just plugging in and see what you get.
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The break down based on the limit laws is the following.
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I just want you to see the breakdown as follows.
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The limit as x approaches 2, the limit of this function.
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This is a quotient, this equals the limit as x approaches 2 of x³ + x² – 4 divided by,
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because the limit of the quotient is the quotient of the limit.
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The limit as x approaches 2 of 6 - 4x.
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This is equal to the limit of the sum is the sum of the limits = the limit as x approaches 2 of x³
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+ the limit as x approaches 2 of x² - the limit as x approaches 2 of 4 divided by the limit as x approaches 2 of 6 - this is 4x.
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I’m going to pull the 4 out, -4 × the limit as x approaches 2(x).
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Now that you have broken down into its limit laws, now we just plug in 2.
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You are going to get 2³ + 2² – 4.
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The limit of the constant is the constant.
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4 × the limit of 2, the limit as x approaches 2(x) is 2.
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It is -4 × 2, you get -4.
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The idea is just put the 2 in, plug it in, see what you get.
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If you get a number, you are done, you can stop.
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If you do not get a number, if you get something that is nonsense, you have to manipulate it.
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We said earlier, as we just said a moment ago, if you plug in and evaluate and you get an actual number, you can stop.
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You are done, this number is your limit, is your answer.
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If when you plug in, if you get something unusual, I will put unusual in quotes.
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If you get something unusual, then you must manipulate f(x) and try the limit again.
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Let us do some examples.
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It will make perfect sense, once you see the examples.
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Let us go ahead, over here, we have got an example.
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We want to calculate the limit as x approaches 1 of x³ + x² - 17x + 15/ x – 1.
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Let us plug in.
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When you put 1 in here, up here is not a problem but you cannot put 1 in here, because 1 - 1 is 0.
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1 is not a domain of this function.
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You get something unusual, we have to manipulate this.
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When we plug in 1, we get 0 in the denominator.
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Let us manipulate by seeing if we can factor this thing.
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Let us see what we can do.
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Let us manipulate this expression, by seeing if we can factor it.
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You have to understand at this point, there is no way for me to know
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what manipulation I'm going to have to do, in order to make this work.
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Factoring is the first thing that I try, simply because I see a rational function.
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I figured just to go ahead and do the long division.
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We will divide the bottom and the top, and we will see what we get.
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That might not work, we might have to try something else.
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There is no way of knowing beforehand.
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There is no single algorithmic procedure that you can follow to solve these limits.
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It is all going to comedown to mathematical ingenuity, mathematical insight, luck, try this and try that.
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Literally, that is what it is going to come down to because things are becoming a lot more complex.
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It is not just a straight single shot where we see the goal and we know that
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we are going to have to take this step to get to that goal.
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If we are faced with another situation, it is not going to be the same step.
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It is never going to be the same steps twice.
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Each problem is individual.
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You have to pull back and get in the habit of not looking to solve a problem immediately, based on what you already know.
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The reason that calculus is actually called analysis is precisely for that reason.
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You have to stop and analyze the situation.
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Take a look at each situation as it arises.
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When you plug in 1, you get a 0 in the denominator, that is unusual, it does not make sense.
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We are going to see if we can simplify this, find an equivalent expression by dividing or seeing if we can factor it.
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Factoring this, it is a cubic equation.
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I’m just going to do the long division.
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That is how I’m going to do it.
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I go ahead and I do x – 1.
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It is going to be x³ + x² – 17x.
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Let us write it so they are legible here.
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My notoriously illegible writing, I apologize for that.
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Let us go ahead and see if we can do this division.
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This is going to be x², x² × x is going to be x³.
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This is going to be - x².
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I'm going to change that sign and change this sign.
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I’m going to cancel that and I'm going to get 2x² - 17x.
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This is going to be +2x, this is going to give me 2x².
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This is going to give me -2x.
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Change that sign, change that sign.
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That cancels and I'm left with -15x + 15.
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This is going to be -15.
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It is going to be -15x + 15.
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Change that sign, change that sign.
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I’m left with 0, perfect.
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Now I get a factorization of x - 1 × x² + 2x – 15.
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This is really great.
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It looks like x² + 2x – 15, it looks like I can actually factor that too.
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That will be x - 1 × x + 5, if I'm not mistaken, × x – 3.
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x² 5x - 3x gives me my +2x, 5 and 3 gives me my -15.
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My top is actually factorable.
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I get, the numerator of the function x³ + x² – 17x + 15 factors.
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We get x - 1 × x + 5 × x - 3/ the denominator which was x – 1.
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That cancels, worked out well.
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We can take the limit again.
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The limit as x approaches 1, not 2.
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Now we plug in 1 to here, we end up with 6 × 1 - 3 is -2 – 12.
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We got our actual answer.
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The original limit is -12.
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In this case, we factor the numerator.
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This is our function, just in factored form of the numerator.
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It turned out that those actually ended up canceling.
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This expression and the original expression are equivalent.
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We take the limit of what is equivalent.
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This time we ended up with a finite number.
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We could stop, our answer is -12.
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We got some nonsense, we fiddled with it, and we came up with an answer.
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Let us try another example here.
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Let us let our f(h), this time our variable will be h.
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Let us let it equal 3 + h² -,
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I’m sorry, 3 + h³ - 27/ h.
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We would like you to find the limit as h approaches 0 of this, of f(h).
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The limit as h approaches 0 of f(h).
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F(h) is this thing.
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It is 3 + h³ - 27/ h.
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When I plug in 0 here, I get 0 in the denominator.
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I cannot do anything with this directly.
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I’m going to have to manipulate it.
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We notice that the numerator is not completely simplified.
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Let us see if we simply it, in other words, expand the 3 + h³.
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Let us see if that turns it into something where we actually can plug in 0 for h and it will give us something that makes sense.
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f(h) is equal to 3 + h³ - 27/ h, that =,
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3 + h³, 1, 3, 3, 1, those are the coefficients of the expansion for an exponent of 3, Pascal’s triangle.
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It is going to be 3³ × 3² h × 3h² h³.
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That takes care of the expansion, that is -27/ h.
00:30:03.900 --> 00:30:19.600
That = 27 + 27h + 9h² + h³ - 27/ h.
00:30:19.600 --> 00:30:23.900
The 27 goes away, here I’m going to factor out an h.
00:30:23.900 --> 00:30:38.900
h × 27 + 9h + h²/ h.
00:30:38.900 --> 00:30:53.600
The h cancels, I'm left with f(h) = 27 + 9h + h².
00:30:53.600 --> 00:30:56.000
This is the same function as we had before.
00:30:56.000 --> 00:30:59.700
It is just simplified and we use algebra to simplify it.
00:30:59.700 --> 00:31:03.000
Now we take the limit again.
00:31:03.000 --> 00:31:16.900
Now we take the limit as h approaches 0 of f(h) which is this.
00:31:16.900 --> 00:31:21.700
27 + 9h + h².
00:31:21.700 --> 00:31:25.100
When we plug in 0, this one goes to 0, this one goes to 0.
00:31:25.100 --> 00:31:29.100
I’m left with an answer of 27, that is my limit.
00:31:29.100 --> 00:31:33.000
That is it, it does not work, simplify it, manipulate it.
00:31:33.000 --> 00:31:36.300
Do whatever you need to do until it works.
00:31:36.300 --> 00:31:41.400
If it does not work the second time, you try it again.
00:31:41.400 --> 00:31:45.900
Welcome to calculus.
00:31:45.900 --> 00:31:49.200
Let us try another example.
00:31:49.200 --> 00:32:03.900
The limit as x approaches 0 of x² + 16 - 3/ x².
00:32:03.900 --> 00:32:12.000
Again, we have a problem that if we plug 0 in, on the top is fine but we are going to have 0 in the denominator.
00:32:12.000 --> 00:32:18.100
We have to do something to this.
00:32:18.100 --> 00:32:37.000
Again, plugging 0 in gives us a 0 in the denom.
00:32:37.000 --> 00:32:42.400
Let us manipulate, this time, we are going to rationalize the numerator.
00:32:42.400 --> 00:32:44.200
You are accustomed to rationalizing the denominator.
00:32:44.200 --> 00:32:45.400
You can do the numerator.
00:32:45.400 --> 00:32:46.300
It actually does not matter.
00:32:46.300 --> 00:32:49.000
You are just going to multiply the top and bottom by the conjugate of the numerator.
00:32:49.000 --> 00:32:51.100
It is that simple.
00:32:51.100 --> 00:33:03.700
Let us manipulate by rationalizing the numerator.
00:33:03.700 --> 00:33:04.600
It is going to look like.
00:33:04.600 --> 00:33:20.000
This is going to be x² + 16, under the radical sign, -3/ x² × x² + 16, all under the radical sign, + 3.
00:33:20.000 --> 00:33:23.300
That is the conjugate of the numerator.
00:33:23.300 --> 00:33:29.100
x² + 16 + 3.
00:33:29.100 --> 00:33:31.200
I’m sorry, did I do this wrong?
00:33:31.200 --> 00:33:39.000
I think this is actually a 4 not a 3.
00:33:39.000 --> 00:33:48.800
This is a 4, this is a 4, and this is a 4.
00:33:48.800 --> 00:33:53.600
I think that is all I have.
00:33:53.600 --> 00:33:57.600
Let us see what we have got, when we actually do the multiplication.
00:33:57.600 --> 00:34:07.500
When we multiply this and this, this and this, we are going to get x² + 16.
00:34:07.500 --> 00:34:10.800
This and this cancel.
00:34:10.800 --> 00:34:22.200
-16/ x² × √x² + 16, under the radical, + 4.
00:34:22.200 --> 00:34:24.900
That and that go away.
00:34:24.900 --> 00:34:37.800
We are left with x²/ x² × √x² + 16 + 4.
00:34:37.800 --> 00:34:47.400
That goes away, we are left with the 1/ x² + 16, under the radical sign, + 4.
00:34:47.400 --> 00:34:49.800
Now we simplify this as much as possible.
00:34:49.800 --> 00:34:51.300
It is the same function.
00:34:51.300 --> 00:34:54.100
All we have done is multiplied by something and change the way it looks.
00:34:54.100 --> 00:34:57.100
We have affected it cosmetically.
00:34:57.100 --> 00:34:59.500
Now we can take the limit again.
00:34:59.500 --> 00:35:11.600
Now the limit as x approaches 0 of this function 1/ x² + 16 + 4.
00:35:11.600 --> 00:35:14.300
We plug 0 in for here, this goes to 0.
00:35:14.300 --> 00:35:25.100
The square root of 16 is 4 = 1/ 4 + 4, our answer is 1/8, an actual number.
00:35:25.100 --> 00:35:29.300
I think you get the idea.
00:35:29.300 --> 00:35:37.100
Let us do some more examples.
00:35:37.100 --> 00:35:44.900
Let us try something a little bit more complicated.
00:35:44.900 --> 00:35:52.400
The limit as x approaches 0 of the absolute value of 3x/ x.
00:35:52.400 --> 00:35:57.200
Absolute values scares the hell out of everyone, including me.
00:35:57.200 --> 00:36:00.600
I just gotten used containing to containing my fear at this point.
00:36:00.600 --> 00:36:04.200
They are daunting, like how do you handle it.
00:36:04.200 --> 00:36:07.500
You remember that the absolute value, it actually consists of two things.
00:36:07.500 --> 00:36:10.800
We have to find two limits here.
00:36:10.800 --> 00:36:15.300
Again, once you put 0 in for x, you are going to get a 0 in the denominator.
00:36:15.300 --> 00:36:16.200
That does not make sense.
00:36:16.200 --> 00:36:20.400
We are going to have to manipulate this somehow.
00:36:20.400 --> 00:36:28.300
The limit of as x approaches 0 of the absolute value of 3x/ x.
00:36:28.300 --> 00:36:33.100
Remember our absolute value sign, the constant inside the absolute value can come out.
00:36:33.100 --> 00:36:43.000
This is actually equal to 3 × the limit as x approaches 0 of the absolute value of x/x.
00:36:43.000 --> 00:36:49.600
We have to deal with this limit actually, and whatever we get we multiply by 3.
00:36:49.600 --> 00:36:52.300
Let us recall what absolute value means.
00:36:52.300 --> 00:36:56.500
The absolute value of x is two things.
00:36:56.500 --> 00:36:59.500
It is equal to x, when x is greater than 0.
00:36:59.500 --> 00:37:08.000
It is equal to –x, when x is less than 0.
00:37:08.000 --> 00:37:16.400
We have to do two separate limits.
00:37:16.400 --> 00:37:29.600
We have to deal with this absolute value of x/x, as two separate limits.
00:37:29.600 --> 00:37:36.800
We have to do one, when x is bigger than 0.
00:37:36.800 --> 00:37:39.500
Remember, we are approaching 0 here.
00:37:39.500 --> 00:37:41.200
0 is what we are approaching.
00:37:41.200 --> 00:37:47.700
When x is bigger than 0, approaching 0 when x is bigger than 0 means we are approaching it from the right.
00:37:47.700 --> 00:37:53.400
It means approaching 0 from there.
00:37:53.400 --> 00:37:57.000
We have to do one for x less than 0.
00:37:57.000 --> 00:38:01.200
We have to do when x approaches 0 on the left.
00:38:01.200 --> 00:38:03.300
We are approaching 0.
00:38:03.300 --> 00:38:08.100
We have to do it this way, that is it from the positive end, from above.
00:38:08.100 --> 00:38:12.600
We have to approach 0 from the negative end, from below.
00:38:12.600 --> 00:38:18.000
Let us see, let me actually change colors here.
00:38:18.000 --> 00:38:22.500
Let me go to purple.
00:38:22.500 --> 00:38:27.900
You know what, purple is nice but I think I like blue better.
00:38:27.900 --> 00:38:31.500
Let us do for x = greater than 0, for that one.
00:38:31.500 --> 00:38:43.000
For x greater than 0, the absolute value of x is x.
00:38:43.000 --> 00:38:50.200
3 × the limit as x approaches 0 of the absolute value of x/x.
00:38:50.200 --> 00:38:54.800
It says the absolute value of x = x for x greater than 0, I just plug in x for here.
00:38:54.800 --> 00:39:02.600
That is equal to 3 × the limit as x goes to 0 of x/x.
00:39:02.600 --> 00:39:12.900
x/x is just 1.
00:39:12.900 --> 00:39:14.400
The limit equals 3.
00:39:14.400 --> 00:39:23.700
If I’m approaching 0 from the right, when x is bigger than 0, my limit of this thing is +3.
00:39:23.700 --> 00:39:35.500
Now let us go x less than 0, let us approach 0 from the negative numbers.
00:39:35.500 --> 00:39:45.400
For x less than 0, the absolute value of x = -x.
00:39:45.400 --> 00:39:53.800
3 × the limit absolute value of x/x = 3 × the limit of x approaches 0.
00:39:53.800 --> 00:40:03.400
Absolute value of x is –x.
00:40:03.400 --> 00:40:10.900
-x/x is -1 = 3 × -1 = -3.
00:40:10.900 --> 00:40:14.200
There you go.
00:40:14.200 --> 00:40:17.200
The limit from the right is 3.
00:40:17.200 --> 00:40:21.100
The limit from the left is -3.
00:40:21.100 --> 00:40:24.800
3 does not equal -3.
00:40:24.800 --> 00:40:33.500
In other words, the right hand limit does not equal the left hand limit.
00:40:33.500 --> 00:40:44.300
This means that the limit as x approaches 0 of the absolute value of 3x/ x does not exist.
00:40:44.300 --> 00:40:46.100
That is how you handle absolute values.
00:40:46.100 --> 00:40:52.100
You actually have to separate it into x being positive and x being negative.
00:40:52.100 --> 00:40:55.100
That is it, I hope that made sense.
00:40:55.100 --> 00:41:07.100
Let us do another example.
00:41:07.100 --> 00:41:19.100
f(x) = combined function, x² – 25, under the radical.
00:41:19.100 --> 00:41:26.200
When x is bigger than 5 and it equals 20 - 4x.
00:41:26.200 --> 00:41:28.900
Sorry, combined function.
00:41:28.900 --> 00:41:35.500
Less than 5 should probably be a little bit more mathematical precise, than I usually am, forgive me.
00:41:35.500 --> 00:41:38.500
We see the 5 is the dividing point.
00:41:38.500 --> 00:41:41.500
When x is bigger than 5, we use this function.
00:41:41.500 --> 00:41:48.100
When x is less than 5, we use this function.
00:41:48.100 --> 00:41:56.800
We want the limit as x approaches 5 of f(x).
00:41:56.800 --> 00:42:00.700
That is what we want, the limit of this function.
00:42:00.700 --> 00:42:03.100
5 is the dividing point.
00:42:03.100 --> 00:42:08.200
Clearly, we are going to have to do an x approaches 5 from above and use this function.
00:42:08.200 --> 00:42:18.800
As an x approaches 5 from below and use this function.
00:42:18.800 --> 00:42:31.600
We evaluate the limit as x approaches 5 from above, by using that function.
00:42:31.600 --> 00:42:36.800
The limit as x approaches 5 from below, by using this function.
00:42:36.800 --> 00:42:40.400
We already know that we have to do the left and right anyway.
00:42:40.400 --> 00:42:42.500
Because it does not specify whether it is left or right.
00:42:42.500 --> 00:42:50.300
We have to actually do left and right, whether they are separate functions or not.
00:42:50.300 --> 00:43:00.800
Let us see what this gives us now.
00:43:00.800 --> 00:43:04.700
Let us do the left hand limit.
00:43:04.700 --> 00:43:12.000
The limit as x approaches 5 from below, we are going to use the one for when x is less than 5.
00:43:12.000 --> 00:43:18.100
Our function is 20 - 4x.
00:43:18.100 --> 00:43:21.100
Just plug 5 in.
00:43:21.100 --> 00:43:25.700
You are approaching it from below, that from below part has nothing to do with the number itself.
00:43:25.700 --> 00:43:27.200
It just means you are approaching it from below.
00:43:27.200 --> 00:43:29.000
You are still approaching 5.
00:43:29.000 --> 00:43:32.600
In order to find out what happens, plug 5 in.
00:43:32.600 --> 00:43:36.500
It is going to be 20 - 20 = 0.
00:43:36.500 --> 00:43:39.500
The left hand limit = 0.
00:43:39.500 --> 00:43:43.400
Let us remind ourselves what f(x) is.
00:43:43.400 --> 00:43:47.600
It is 20 - 4x and it is x² – 25.
00:43:47.600 --> 00:43:49.400
This is for when x is greater than 5.
00:43:49.400 --> 00:43:52.400
This is for when x is less than 5.
00:43:52.400 --> 00:43:59.900
Now the right hand limit, the limit as x approaches 5 from above, this function.
00:43:59.900 --> 00:44:03.900
It is going to be x² – 25.
00:44:03.900 --> 00:44:11.700
Plug it in, you are going to get 25 - 25 under the radical = 0.
00:44:11.700 --> 00:44:15.900
Here the left hand limit = the right hand limit.
00:44:15.900 --> 00:44:43.500
Therefore, which implies that the limit exists and the limit approaches 5 of f(x) = 0.
00:44:43.500 --> 00:44:52.600
Let us write down a couple of theorems that may actually help in the evaluation of limits.
00:44:52.600 --> 00:45:00.700
Some theorems that may help.
00:45:00.700 --> 00:45:27.400
The first theorem, if f(x) is less than or equal to g(x) near a point a and the limit as x approaches a(f)
00:45:27.400 --> 00:45:45.800
and the limit as x approaches a(g) both exist, then the limit as x approaches a of f(x)
00:45:45.800 --> 00:45:55.500
is less than or equal to the limit as x approaches a of g(x).
00:45:55.500 --> 00:46:00.000
Basically, f(x) is less than or equal to g(x), you already know that what you do to the left side,
00:46:00.000 --> 00:46:09.000
if you do it to the right side, any operation that you take, retains the relation.
00:46:09.000 --> 00:46:12.900
If I multiply f(x) by 5, I multiply g(x) by 5.
00:46:12.900 --> 00:46:20.100
5 f(x) is less than or equal to 5 g(x), because f(x) is less than g(x).
00:46:20.100 --> 00:46:21.000
It is the same thing.
00:46:21.000 --> 00:46:22.500
f(x) is less than or equal to g(x).
00:46:22.500 --> 00:46:33.000
Therefore, the limit of f(x) is less than or equal to the limit of g(x), provided both limits exist.
00:46:33.000 --> 00:46:38.800
The next one which is pretty important.
00:46:38.800 --> 00:46:41.800
Probably we are going to use it, but every once in while it might come up.
00:46:41.800 --> 00:46:44.900
It is among that is hardest to remember.
00:46:44.900 --> 00:46:49.700
I think in my entire mathematical career, I think I have used it 4 times.
00:46:49.700 --> 00:47:06.800
If f(x) is less than or equal to g(x), it is less than or equal to h(x) and
00:47:06.800 --> 00:47:22.700
the limit as x approaches a of f(x) = the limit as x approaches a of h(x), which happens to equal a,
00:47:22.700 --> 00:47:32.500
if this relation exists and the limit of f(x) and the limit of f(x), the two flanking functions,
00:47:32.500 --> 00:47:44.500
if they happen to have the same limit then the limit as x approaches a of g(x) also = a.
00:47:44.500 --> 00:47:49.700
It makes sense, if the limit of this is 5 and the limit of this is 5, this is in between those two.
00:47:49.700 --> 00:47:53.900
The limit has to be 5, that is pretty much what is going on.
00:47:53.900 --> 00:48:06.800
It is called the squeeze theorem.
00:48:06.800 --> 00:48:11.600
Graphically, it looks like this.
00:48:11.600 --> 00:48:16.400
Let us say this is our point a, let us say it is over there.
00:48:16.400 --> 00:48:20.000
Let us say this is our a.
00:48:20.000 --> 00:48:28.200
Let us say we have some function which is something like that.
00:48:28.200 --> 00:48:36.000
Then, maybe something like that and something like this.
00:48:36.000 --> 00:48:40.200
Let us let this be the h(x).
00:48:40.200 --> 00:48:47.000
Let this be the g(x) and let this be f(x), this is a.
00:48:47.000 --> 00:49:00.300
Near a, you see that f(x) is less than g(x) is less than h(x).
00:49:00.300 --> 00:49:10.800
As you get close to a, if the limit of f(x) is a, the limit of h(x) is a,
00:49:10.800 --> 00:49:15.300
basically g has no choice but to be squeezed in between them.
00:49:15.300 --> 00:49:23.100
The limit of g(x) is equal to a, that is why they call it the squeeze theorem.
00:49:23.100 --> 00:49:33.300
Let us go ahead and do an example of one of these.
00:49:33.300 --> 00:49:43.300
What is the limit as x approaches 0 of 5x² × cos(1/x).
00:49:43.300 --> 00:49:47.800
Clearly, if we plug 0 in, we cannot because we have 1/0 here.
00:49:47.800 --> 00:50:08.200
That is not going to work, we have to do something.
00:50:08.200 --> 00:50:19.300
We do know one thing, we know something about cosine.
00:50:19.300 --> 00:50:22.600
This fact about sine and cosine, very important fact.
00:50:22.600 --> 00:50:26.800
Remember this one fact, it will probably save you a lot of grief and
00:50:26.800 --> 00:50:31.600
make a lot of problems that are otherwise intractable, very easy to solve.
00:50:31.600 --> 00:50:37.000
We know something about cos(1/x).
00:50:37.000 --> 00:50:42.400
In other words, the cos(a) whatever a happens to be, some function of x.
00:50:42.400 --> 00:50:50.600
We know that the cos(1/x) lies between 1 and -1.
00:50:50.600 --> 00:50:57.500
The sine and the cosine functions, they maximum value is 1 and their minimum value is 1, always.
00:50:57.500 --> 00:50:59.600
We know that this is true.
00:50:59.600 --> 00:51:02.600
Since that is the case, watch this.
00:51:02.600 --> 00:51:04.400
This is a relationship that is true.
00:51:04.400 --> 00:51:23.000
I’m going to multiply everything by 5x² which means 5x² × -1 is less than or equal to 5x² × cos(1/x) is less than or equal to 5x² × 1.
00:51:23.000 --> 00:51:32.000
This is just -5x², this is 5x² × cos(1/x) which is our original function.
00:51:32.000 --> 00:51:36.200
It is less than or equal to 5x².
00:51:36.200 --> 00:51:39.200
Let us take the limit of this and this function, and see what we get.
00:51:39.200 --> 00:51:43.700
Again, this is true, limit, limit, limit.
00:51:43.700 --> 00:51:54.900
If I apply the same operation to everything in this relational chain, the relation is retained.
00:51:54.900 --> 00:52:08.100
The limit as x approaches 0 of -5x² is going to be less than or equal to the limit as x approaches 0 of 5x²
00:52:08.100 --> 00:52:16.800
× the cos(1/x) is going to be less than or equal to the limit as x approaches 0 of 5x².
00:52:16.800 --> 00:52:30.300
Plugging here, we get 0 is less than or equal to the limit as x approaches 0 of 5x² × cos(1/x),
00:52:30.300 --> 00:52:32.100
less than or equal to this limit.
00:52:32.100 --> 00:52:33.900
When we plug it in, we get 0.
00:52:33.900 --> 00:52:45.600
0,0, therefore, our original limit is 0.
00:52:45.600 --> 00:52:51.900
You are more than welcome to graph it yourself, to actually see that it is 0.
00:52:51.900 --> 00:52:58.300
There you go, that is calculating limits mathematically, calculating them analytically.
00:52:58.300 --> 00:53:02.200
Plug in the value that x is approaching and see what happens.
00:53:02.200 --> 00:53:03.400
If you get a number, you can stop.
00:53:03.400 --> 00:53:04.700
You are done, that is your limit.
00:53:04.700 --> 00:53:08.600
If not, you are going to have to subject the function to some sort of manipulation.
00:53:08.600 --> 00:53:10.100
You are going to be converting it into something equivalent.
00:53:10.100 --> 00:53:11.900
You are not going to changing it.
00:53:11.900 --> 00:53:17.000
You are converting it to something equivalent using the various tools that you have at your disposal.
00:53:17.000 --> 00:53:23.300
Factoring, the squeeze theorem, rationalizing numerators, rationalizing denominators,
00:53:23.300 --> 00:53:29.300
whatever else that your own peculiar personal ingenuity can come up with.
00:53:29.300 --> 00:53:31.700
That is the wonderful thing about these, is every year,
00:53:31.700 --> 00:53:37.200
it amazes me the different ways that kids come up with solving this limits.
00:53:37.200 --> 00:53:44.400
I mean it is almost infinite, the number of variations that they can come up with, it is exciting.
00:53:44.400 --> 00:53:46.500
In any case, thank you for joining us here at www.educator.com.
00:53:46.500 --> 00:53:48.000
We will see you next time, bye.