WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about finding limits.
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We often need to find the precise value that a limit will produce.
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However, the methods we saw when we first introduced limits (that is, graphing it or a table of values for the function) are not precise.
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They can give us a good idea of what the limit will be, but they don't give us certainty.
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They don't let us know that it will be exactly something.
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Likewise, the formal (ε,Δ) definition of a limit that we talked about in the last lesson--
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and it is totally fine if you do not know it; that was a completely optional lesson, only if you are really interested in math,
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and wanted to find out more about stuff that is going to come later in a few years if you keep studying math;
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it is totally fine if you didn't do it--but if you did, that is still not really going to help us find limits.
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It allows us to formally prove that this limit has to be here.
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And it is the deeper mechanics of what is going on "under the hood" for how a limit works.
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But it doesn't let us find limits; it doesn't make finding them easier; it is just about proving limits.
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In this lesson, we will see various methods to find the precise value; that is what this lesson will be about.
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The basic idea that we are going to see with all of these methods is to transform the function into something
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that works pretty much the exact same way, that we will just be able to plug in a value for the x
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in this equivalent version, and we will be able to churn out some value for what the limit will come out to be.
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Now, before you watch this, make sure that you are already familiar with the concept of a limit.
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You really want to have a good understanding of how a limit works.
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We will be working out how to get numbers in this lesson.
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But if you don't actually understand what this stuff means, it is all going to fall apart really quickly.
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So, it is really important that you understand how a limit works before you watch this.
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If you don't already have a good understanding of how limits work, check out the lesson two lessons back,
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Idea of a Limit, where we will explain and get an idea of what a limit is about.
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And that way, we will have some meaning for how we actually get precise values.
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You can figure out the precise values without really understanding what is going on.
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But that will fall apart really, really quickly; and you might as well just have a nice foundation to work from there.
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It won't take that long to get an idea of what is going on.
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All right, first, the easiest limits to find are limits for "normal functions."
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And now, that is in quotes, because normal is not a technical term.
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What I just mean here is...it is supposed to mean the kind of functions that we are used to dealing with,
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the sort of thing that we use most often, functions where it does not break at the point we are interested in--
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that is to say, the function is defined and makes sense; it doesn't suddenly break down when we get to the place that we really care about.
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And it is not piecewise, and/or the point that we are interested in is not at the very edge of the domain.
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The point that we care about, whatever x we are going to (we are going to some x going to c)...
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whatever c we are going towards, it is not going to be at the very edge of the domain, or where we split on some piecewise function.
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So, as long as that location makes sense--everything in that area makes sense--
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we know how to use the function in that area around that place, and it isn't piecewise--
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there aren't different parts of it, and it is not the very edge of the domain,
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the very starting or very last value for it--as long as those aren't the case, we will be able to do it really easily.
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If these two conditions are met, where it isn't breaking down and it is not piecewise,
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and the point isn't at the edge of the domain, then it is really easy to figure out what the value is going to be.
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So, if the point we are interested in is the value that x approaches in the limit--
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that is to say, x goes to c, and c is the place where things aren't breaking down,
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and c is not at a piecewise break-over, and it is not at the very edge of the domain,
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then it is almost always going to be the case that the limit as x goes to c of f(x)
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is as simple as just plugging c in for x and getting f(c).
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So, let's get an example first, to see how we could use this.
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If we looked at the limit as x goes to 2 of 1/x², what would that end up being?
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Well, first notice: while 1/x² breaks down (it isn't defined, that is to say) at x = 0, that is here.
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But we don't care about x = 0; that is not the area that we are interested in.
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We care about x going to 2; so if x is going to 2, if we look in that area over here, that region, that is totally fine; it makes perfect sense.
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1/x² works fine in the region around x going to 2.
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As x goes to 2, well, all of this stuff makes sense; we can see it.
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It clearly just maps to values, and it is perfectly reasonable.
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Second, 1/x² is not a piecewise function; we don't have to worry about that.
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And x = 2 is not at the edge of the domain (and the domain for 1/x² is every x, with the exception of x equaling 0; it is all x not equal to 0).
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So, we are not at the edge of a domain; it is not piecewise; and it makes sense in the area we are looking for.
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It makes total sense in here; so since it does all of those, it means that we can just plug in the value that we are going to.
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Because of these two conditions, limit as x goes to 2, we just plug it in for x,
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and we have 1/2², which simplifies to 1/4, and there is our limit.
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Why can we do this--what is the reason that we can get away with doing this?
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Well, in general, most of the functions we are used to working with don't do anything weird.
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They aren't strange in any way; and what I mean by not being weird is that they are defined everywhere; they don't have holes; they don't jump around.
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They are defined anywhere that we might be interested in looking; they don't have any holes in them; and they don't jump around.
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They work in a pretty reasonable way; they work normally--they are not weird.
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What this all means is that the functions we are used to working with, the functions that we normally work with, go where we expect them to go.
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Since a limit is about figuring out where a function is headed (a limit is about what our expectation is
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for this function), and a normal function has our expectations fulfilled (what we expect from the function
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is what we get out of the function), that means that we can evaluate normal functions at the location the limit approaches.
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Our expectation, the limit as x goes to c of f(x), what we expect to end up landing on in our journey,
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what we expect as we come in, ends up being what it actually is--what it is at the location.
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So, if it is normal, if f(x) isn't doing something weird, then what we expect ends up being what we actually get.
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So, the limit as x goes to c...well, we can just plug that in for f(x),
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and we have that f(c) will end up being the limit, any time that we are dealing with a normal function.
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In fact, this is true even if f(x) does have weird stuff, but as long as it happens somewhere else.
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All we care about is x going to c; so as long as the neighborhood around x going to c is normal,
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and the weird stuff happens off somewhere else--it isn't happening directly on top of that c--
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then the weird stuff...we don't care about it, because the region we care about being normal is the region around x = c.
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As long as the neighborhood around x = c is normal, as long as around x = c does this,
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then we will end up being able to just plug into that, just fine.
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So, if there is weird stuff, that can be OK, as long as it is far enough away.
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When we looked at 1/x², there was weird stuff at x = 0, but we didn't care about x = 0.
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We cared about x going to 2; so as long as the weird stuff isn't right on top of where we are going to,
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we can just plug in our c, and that will tell us what the limit will come out to be,
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if it is this fairly normal function that we have been working with for years and years.
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Of course, sometimes x goes to c, but something weird does happen at x going to c.
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When we are at x = c, it is a weird place.
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One weird thing that often happens is dividing by 0; you are not defined when you divide by 0.
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So, consider the function and limit below: f(x) = 2x/x.
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Well, we can see it graphed here; it makes perfect sense, up until we try to plug in x = 0, at which point the function breaks down.
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But the limit is pretty clear; it is going to 2 the entire time, so what is it headed towards?
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It is headed towards 2; that is what we get out of the limit.
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So, of course, we can clearly see that the limit in this case exists, and it is 2.
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However, let's also notice that with the exception of x = 0, everywhere other than actually at x = 0,
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where the weird thing happens, the function f(x) = 2x/x is just equivalent to if we had canceled out those x's and gotten g(x) = 2.
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Notice: f(x) = 2x/x and g(x) = 2...these two functions are identical, with the exception at x = 0.
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Everywhere other than x = 0, these two are totally the same.
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f(x) = 2x/x and g(x) = 2 behave exactly the same, with an exception at x = 0.
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However, since we are looking at the limit as x goes to 0, we don't care about x = 0.
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It is about the journey, not the destination; so if it is x to 0, the destination we don't care about is at x = 0.
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So, the weird thing here at x = 0...we might as well forget about it.
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That means, since we don't care about what happens at the weird place, and g(x) = 2 is exactly the same
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everywhere but the weird place, that means that we can use g(x) to evaluate the limit.
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And we can just plug in g(0) to get 2.
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So, g in general works just the same as f; g is just the same as f.
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It works the same everywhere, with the exception of this one weird little point.
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However, since we are looking at a limit going to that one weird little point, we don't actually care about the weird little point.
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A single point doesn't matter, because the limit is about the journey towards that point.
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So, g(x) = 2 behaves the exact same for the journey portion.
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The journey towards behaves the exact same, whether we are using f or g, which means that g(x) is what we can use for figuring out the limit.
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And then, by the same logic we were just talking about previously, since g(x) is totally normal at 2,
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we can just plug in there, and we can get the answer from g(x).
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Now, how did we find g(x)?--by canceling common factors.
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This logic works in general; if we have some function given as a fraction, we can cancel out factors between top and bottom,
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because a single point...if we cancel a factor, the only thing that can possibly happen
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that would be bad is that we would cause one point to change around, like with f(x), x = 0 did not exist;
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but with g(x), x = 0 did exist; so we caused a problem for specifically one point.
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But a single point has no effect on a limit, because with a limit, it is about the journey, not a single point that is our destination.
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So, since a journey is made up of a multitude of points, taking out a single point,
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changing a single point, doesn't actually have an effect on where we end up landing for our limit.
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That means, any time we have common factors for a limit, we can cancel out common factors
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and just get what it would be without those common factors for the limit.
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We will have changed the function that we are using, but the limits will be equivalent.
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So, here is an example: we have limit as x goes to 3 of (x - 3)(x + 2)/(x - 3).
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Well, that means that, if we were to plug x = 3 into this, we would have 0/0; 3 - 3 turns to 0; 3 - 3 on the bottom turns to 0; so we would get 0/0.
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And so, we can't do that; it goes crazy; it is weird there.
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But we can cancel out x - 3 and cancel out x - 3, just like we canceled out the k's here; and we get some other something over something.
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We get what remained, a and b; in this case, we have x + 2 divided by 1 now.
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So now it is the limit as x goes to 3 of x + 2; that is effectively going to work the same.
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x + 2 is pretty much equivalent to x - 3 times x + 2 over x - 3, with the exception of x = 3.
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But since we don't care about that, since that is where we are headed towards, we can end up using that limit instead.
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So, we now plug in x going to 3 into x + 2.
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Well, x + 2 is perfectly normal at x = 3; nothing weird happens there.
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So, since nothing weird happens there, it works normally; we can just plug our value in there.
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We plug in 3; 3 + 2...we get 5; our answer is 5 for the limit.
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A similar idea to canceling out factors is rationalization: rationalizing a portion of some fraction.
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If we have a radical that is in our way, and it is part of a fraction, we can change it into a non-radical
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by multiplying the numerator and denominator by the same thing.
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What do we get rid of a radical with--how do we rationalize an expression?
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We rationalize an expression that contains a radical by multiplying by its conjugate.
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That is the same expression, but now with a negative on one side.
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For example, if we have √x + 2, well, that is a radical and some non-radical thing.
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If we go through the conjugate process, we get √ - 2; plus switches to negative.
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If we have √(x² - 3x) - 47x, some radical minus something that is not in a radical,
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then its conjugate is √(x² - 3x) + 47x.
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So notice: plus, if we are going through a conjugate, becomes minus; and minus becomes plus.
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We just swap the sign on one of the portions, and that is how we get the conjugate.
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Now, if we multiply a radical expression by its conjugate, we will end up canceling out the radicals.
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And we will see how that works in just a moment.
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Since multiplying a fraction on the top and bottom by the same thing gives an equivalent expression--
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if I have a fraction, I can multiply it by 5/5, because 5/5 is just the same thing as 1, so it is still equivalent--
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we can figure out limits by doing this thing.
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And we can trust that this works, that this doesn't introduce any issues, by the exact same logic that we use to cancel out factors.
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If we were to multiply by something over something, it could introduce one slight issue;
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but it would only introduce a slight change to the function at one point.
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But because it is a limit, we don't care about single points on their own; we only care about continuums of points.
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So, that single point being changed is not really an issue; the limit will still work the same.
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So, if we have the limit as √(x + 4) - 2, over x, as x goes to 0,
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well, we would want to multiply this by the conjugate, √(x + 4)...but it was a minus previously,
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so now it swaps to a + 2, and it will have to be divided by the same thing,
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because we can only multiply by 1 effectively, which means the same thing on top and bottom: + 2.
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And we multiply by that, and we start working things out.
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What do we get out of that? Well, limit as x goes to 0 of √(x + 4) - 2, over x, times √(x + 4) + 2,
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over √(x + 4) + 2...well, the top here is now going to become x + 4 - 4.
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And if we are not quite sure how we see that, let's look at √(x + 4) - 2 times √(x + 4) - 2.
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Well, what is...oops, I should not have put that radical over the whole thing; the radical ends there.
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What is √(x + 4) times √(x + 4)?
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Well, the square root of thing times the square root of same thing always just lets out the thing on its own.
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The square root of smiley-face times the square root of smiley-face becomes smiley-face.
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So, √(x + 4) times √(x + 4) becomes x + 4.
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So, x + 4 is what we get out of that; and now we have √(x + 4) times -2...
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oops, that shouldn't be -2; it should be plus, because it is a conjugate; I'm sorry about that.
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√(x + 4) times positive 2 is 2√(x + 4): and then -2 times √(x + 4) is -2√(x + 4).
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So, we have positive 2 √(x + 4) and -2√(x + 4); those two things cancel each other out.
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Positive 2 and negative 2 cancel each other out; so the middle part disappears.
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And now, it is -2 times +2; that becomes -4; so that is where we get x + 4 - 4.
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On the bottom, it is x times this thing over here; so we just put in the quantity, because we will have some convenient canceling happening very soon.
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On our top, we have x + 4 - 4; so plus 4 and minus 4 cancel each other out; we are left with just x on top;
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it is still x times √(x + 4) + 2 on the bottom.
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But now, we say, "We have x on top and x on bottom," and now we have the limit as x goes to 0 of 1/(radic;(x + 4) + 2).
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At this point, we see that if we plug in 0...does anything weird happen?
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Well, 1/(√(0 + 4) + 2)...we aren't dividing by 0 anymore, so it effectively works as a normal function.
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We don't have any weird thing happening, so we plug in for our x at this point.
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The square root of 0 + 4, plus 2 is in our denominator: 1 over...√4 is 2, so we have 1 over 2 + 2, which gets us 1/4.
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And that is how we get to our answer for that limit.
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We will discuss evaluating the limits of piecewise functions.
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We haven't talked about piecewise functions yet, because we will be talking about that in the next lesson, Continuity and One-Sided Limits.
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We will want a couple of little new ideas before we talk about piecewise functions.
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That is why we are saving them for the next lesson.
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For now, though, remember: as long as you are not trying to evaluate--as long as it is not trying
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to evaluate the limit at some piecewise break-over, where it switches from one piece to another piece,
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the function is probably going to be behaving normally on the pieces that contain the point you care about.
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It might have a piecewise here and a piecewise here, and then it suddenly switches over.
00:17:01.500 --> 00:17:06.700
But as long as we are over here in the normal area, or we are over here in the normal area, nothing weird happens.
00:17:06.700 --> 00:17:12.000
So, if it is not at a break-over point, then that means that, since it is behaving normally,
00:17:12.000 --> 00:17:16.800
you can approach it the same way as you do when just dealing with the limit of a normal function.
00:17:16.800 --> 00:17:20.500
Just plug in the appropriate value and see what comes out.
00:17:20.500 --> 00:17:23.300
Plug in the value that you are going towards and see what comes out.
00:17:23.300 --> 00:17:28.000
However, if you do need to evaluate a break-over point, where you have to be talking
00:17:28.000 --> 00:17:31.500
about where it switches from one to the next, check out the next lesson,
00:17:31.500 --> 00:17:34.800
because we will see how that idea works specifically in the next lesson.
00:17:34.800 --> 00:17:36.700
All right, we are ready for some examples.
00:17:36.700 --> 00:17:41.700
The first one: Evaluate the limit as x goes to 2 of x⁴ - 3x² + 4x - 10.
00:17:41.700 --> 00:17:46.500
Our first question that we want to ask ourselves is if x⁴ - 3x² + 4x - 10 is normal.
00:17:46.500 --> 00:17:50.700
Yes, there is nothing weird that happened in that; that is just a polynomial--we are used to that sort of thing.
00:17:50.700 --> 00:17:57.900
So, it is normal; if it is normal, then that means that we can just plug in our value for each of the x.
00:17:57.900 --> 00:18:04.400
So, we plug in our 2, because we know that the limit of what it gets is...well, we don't have to plug x into 10...
00:18:04.400 --> 00:18:09.300
we know that the limit as what we are going to get out of this is just the same thing as what the function would be there.
00:18:09.300 --> 00:18:13.800
What we expect is what we get when we are dealing with a normal function.
00:18:13.800 --> 00:18:23.300
So, we plug in 2; we now have 2⁴ - 3(2)² + 4(2) - 10.
00:18:23.300 --> 00:18:29.000
2⁴ is 16; minus...3(2)²...2² is 4; 3 times 4 is 12,
00:18:29.000 --> 00:18:40.300
plus...4 times 2 is 8; minus 10; 16 + 8 gets us 24; minus 12 minus 10 gets us -22; we put those together, and we now have 2--done!
00:18:40.300 --> 00:18:47.600
The next one: let f(x) equal x² - 3 when x is less than or equal to 2 and 5x + 2 when x is greater than 2.
00:18:47.600 --> 00:18:50.800
First, let's just see, really quickly, what this looks like.
00:18:50.800 --> 00:18:59.100
Here is a rough picture of what it looks like; I will do that with blue here; so x² - 3...that is basically like a parabola.
00:18:59.100 --> 00:19:02.800
It has just been lowered by 3 from a normal standard parabola.
00:19:02.800 --> 00:19:06.800
And it goes until x ≤ 2, at which point it stops here, at 2.
00:19:06.800 --> 00:19:12.700
And then, after that, we are at x > 2, so we switch over to 5x + 2 when x is greater than 2.
00:19:12.700 --> 00:19:15.400
It starts here, and then it goes off like this.
00:19:15.400 --> 00:19:17.600
And that is what we end up seeing.
00:19:17.600 --> 00:19:25.100
The question here is: if we are going to evaluate the limit as x goes to 1 of f(x)...oh, no, it is a piecewise function!
00:19:25.100 --> 00:19:29.100
Well, yes; but we are clearly contained within x ≤ 2.
00:19:29.100 --> 00:19:33.700
The area we care about is this area here; that is far enough from something weird happening.
00:19:33.700 --> 00:19:38.300
Something weird does end up happening over a little bit further to the right.
00:19:38.300 --> 00:19:43.300
But we don't care about that, because in the neighborhood we are interested in (that is x going to 1),
00:19:43.300 --> 00:19:47.600
we are definitely less than or equal to 2, if we are close enough to 1.
00:19:47.600 --> 00:19:54.500
So, since being close enough to 1 means nothing weird happens, all we are dealing with is x² - 3.
00:19:54.500 --> 00:20:01.500
So, we are effectively normal, because we only have to consider the part of the piecewise function that we are inside of.
00:20:01.500 --> 00:20:04.500
The part of the piecewise function that we are inside of is x² - 3.
00:20:04.500 --> 00:20:15.700
So, we can just plug our 1 into x² - 3; we plug in our 1; 1² - 3; 1 - 3 gets us -2, and there is our limit.
00:20:15.700 --> 00:20:20.900
The next example: Evaluate the limit as x goes to -3 of (x² + 3x)/(x² - 9).
00:20:20.900 --> 00:20:25.900
Our first question is, "Is it normal?" Well, if we plug in -3, what do we get on the top?
00:20:25.900 --> 00:20:30.700
Well, we will get 9 - 9; so that is 0; and on the bottom, we will get x² - 9,
00:20:30.700 --> 00:20:37.500
so that will be positive 9, minus 9; oh, we get 0/0; so that means it is not normal.
00:20:37.500 --> 00:20:40.700
Oh, no! But what do we do as soon as we are not normal?
00:20:40.700 --> 00:20:43.400
We start thinking, "All right, well, what else can we do?"
00:20:43.400 --> 00:20:51.200
We have the possibility of canceling factors; so what we want to do now is think, "Is there a way for us to cancel out factors?"
00:20:51.200 --> 00:20:54.400
Can we cancel out factors? Well, we have x² + 3x...
00:20:54.400 --> 00:20:59.300
so I will write limit as x goes to -3...technically, we really should have the limit at every step.
00:20:59.300 --> 00:21:02.700
If you end up really not feeling like writing out the whole thing, at least write limit, so that we know
00:21:02.700 --> 00:21:05.900
that we are still dealing with the limit; and we will plug in something later.
00:21:05.900 --> 00:21:16.700
(x² + 3x)/(x² - 9): well, limit as x goes to -3...how can we change the top?
00:21:16.700 --> 00:21:22.500
How can we factor the top? Well, we could pull out an x, and we would have x(x + 3) on top.
00:21:22.500 --> 00:21:29.600
How can we factor the bottom, x² - 9? Oh, that is just the same thing as (x + 3)(x - 3).
00:21:29.600 --> 00:21:35.500
Great; we can cancel factors; we cancel x + 3 and cancel x + 3.
00:21:35.500 --> 00:21:38.100
And that is fine, because we are just working with a limit.
00:21:38.100 --> 00:21:47.100
Our limit is now the same thing as x goes to -3 of x on top, divided by x - 3 on the bottom.
00:21:47.100 --> 00:21:51.500
Now, we ask ourselves, "If we plug in -3, does anything weird and disastrous happen?"
00:21:51.500 --> 00:21:57.000
-3 on top; -3 on the bottom; we don't get 0/0; we don't even get dividing by 0 once.
00:21:57.000 --> 00:22:03.800
We are totally fine; so we plug in, because now it is effectively a normal function, since something weird isn't happening.
00:22:03.800 --> 00:22:17.600
So, we have -3/(-3 - 3); that gets us -3/-6; the negatives cancel; 3/6 is 1/2, and so we get 1/2 as the answer to our limit.
00:22:17.600 --> 00:22:22.400
Great; the next one: Evaluate the limit as x goes to 0 of tan(x)/sin(x).
00:22:22.400 --> 00:22:26.800
The first question that we ask ourselves is, "Is this normal?"
00:22:26.800 --> 00:22:37.300
Well, if we plug in 0, tan(0) is 0; sin(0) is 0; oh, that means it is 0/0; so it is not normal.
00:22:37.300 --> 00:22:43.200
But if it is not normal, the first thing we try is asking ourselves, "Can we cancel factors?"
00:22:43.200 --> 00:22:47.600
So, if we can cancel factors, we are good; how can we cancel factors?
00:22:47.600 --> 00:22:52.400
Limit as x goes to 0...how can we change tan(x)?
00:22:52.400 --> 00:22:56.700
Well, tan(x)...remember, that is just the same thing as sin(x)/cos(x).
00:22:56.700 --> 00:23:00.000
And any time we don't have just sine and cosine, and we are dealing with trigonometric stuff,
00:23:00.000 --> 00:23:09.900
it normally helps to put it just in terms of sine and cosine; so we have sin(x)/cos(x), all divided by sin(x).
00:23:09.900 --> 00:23:14.700
Oh, OK; well, now we see that we can start canceling stuff; limit as x goes to 0...
00:23:14.700 --> 00:23:20.900
well, we could rewrite this as sin(x)/cos(x); we could just cancel out the sin(x) if we see that directly.
00:23:20.900 --> 00:23:26.700
But if we find it difficult to divide fractions and fractions, we could think of this as divided by sin(x),
00:23:26.700 --> 00:23:36.000
which means that that is the same thing as limit as x goes to 0 of sin(x)/cos(x), times 1/sin(x).
00:23:36.000 --> 00:23:42.500
If you divide, it flips to a fraction, which we could have also done by just breaking out the fraction of the sin(x) on the bottom to the side.
00:23:42.500 --> 00:23:45.800
So now, we see that there is sin(x) on the bottom and sin(x) on the top.
00:23:45.800 --> 00:23:52.600
We cancel some stuff out; limit as x goes to 0...now we have 1/cos(x).
00:23:52.600 --> 00:23:58.400
Since we managed to cancel some stuff out, let's ask ourselves, "If we plug in 0, is it weird? Is it normal? What happens?"
00:23:58.400 --> 00:24:04.800
Well, x goes into 0 for cos(x); cos(0) is 1; 1/1 is totally not weird anymore.
00:24:04.800 --> 00:24:20.100
So, that means we can now swap in our 0; 1/cos(0)...cos(0) is 1; we have 1/1, so our limit comes out to be 1.
00:24:20.100 --> 00:24:25.700
OK, the next one: Evaluate the limit as x goes to 4 of (2 - √x)/(4 - x).
00:24:25.700 --> 00:24:28.300
The first thing we ask ourselves is, "Is it normal?"
00:24:28.300 --> 00:24:32.300
Well, if we plug in 4 on the top, we will get 2 - √4, so that will come out to be 0,
00:24:32.300 --> 00:24:40.600
divided by 4 - 4...even worse...dividing by 0...so 0/0 is not normal--no! No!
00:24:40.600 --> 00:24:44.500
It is not normal, so the next thing we ask ourselves is if we can cancel factors.
00:24:44.500 --> 00:24:50.400
Well, 2 - √x...4 - x...we might be able to figure out a way to cancel factors, but not easily.
00:24:50.400 --> 00:24:55.700
Canceling factors...we might be able to figure out a way; we could figure out a way;
00:24:55.700 --> 00:25:00.700
but let's say that we don't want to figure out canceling factors; canceling factors is not easy.
00:25:00.700 --> 00:25:05.300
So, there is a radical; what was the trick we learned for dealing with radicals?
00:25:05.300 --> 00:25:13.700
We rationalize; we move on to the next trick in our selection; we rationalize.
00:25:13.700 --> 00:25:23.100
What do we do? We have the limit as x goes to 4 of 2 - √x, over 4 - x.
00:25:23.100 --> 00:25:26.200
How do we rationalize? We multiply by the conjugate on the top and the bottom.
00:25:26.200 --> 00:25:34.600
2 - √x: its conjugate is 2 + √x; we could also put a negative on the 2, but it doesn't really matter which side gets the negative.
00:25:34.600 --> 00:25:42.300
2 + √x; 2 + √x; great--multiply our tops together, and multiply our bottoms together.
00:25:42.300 --> 00:25:50.600
2 - √x times 2 + √x: these are now factors with parentheses around them, because we have to have distribution going on.
00:25:50.600 --> 00:25:58.800
So, 2 times 2 is 4; 2 times √x is 2√x; -√x times 2 is +2√x and -2√x; they cancel each other out;
00:25:58.800 --> 00:26:05.000
-√x times +√x becomes -x (√x times √x always comes out to be x;
00:26:05.000 --> 00:26:09.400
√smiley-face times √smiley-face always comes out to be smiley-face; root(root) cancels the roots,
00:26:09.400 --> 00:26:11.500
as long as it is the same thing underneath it).
00:26:11.500 --> 00:26:17.300
4 - x times 2 + √x: well, we could multiply them together--but 4 - x is what we have on the top right now.
00:26:17.300 --> 00:26:21.900
We are basically working towards canceling out factors; so 2 + √x...
00:26:21.900 --> 00:26:26.800
at this point, we are now going to be able to cancel out factors.
00:26:26.800 --> 00:26:32.500
We have the 4 - x on the bottom and the 4 - x on the top; they cancel each other out.
00:26:32.500 --> 00:26:41.200
And now, I have the limit as x goes to 4 of 1/(2 + √x).
00:26:41.200 --> 00:26:44.900
Great; so, if we were to plug 4 into this, would something weird happen?
00:26:44.900 --> 00:26:49.300
1/(2 + √4)...no: we don't have 0 showing up; we aren't dividing by 0.
00:26:49.300 --> 00:26:52.800
It basically works like a normal function; it does work like a normal function.
00:26:52.800 --> 00:26:56.900
Nothing weird is going on there; so that means we can just plug in our value.
00:26:56.900 --> 00:27:11.700
So, 1 over 2 plus...plugging in 4 for our x...1 over 2 + 2 (the square root of 4 is 2)...and 1/4 is our answer; nice.
00:27:11.700 --> 00:27:18.500
The final example: Evaluate the limit as x goes to 0 of 1/(x + 3) - 1/3, all divided by x.
00:27:18.500 --> 00:27:21.500
The first question we ask ourselves is, "Is it a normal function?"
00:27:21.500 --> 00:27:26.300
So, is it normal? Well...you can guess by my hint: no.
00:27:26.300 --> 00:27:34.400
If we plugged in 0, it would be 1/3 - 1/3, so 0 on top, divided by 0 on bottom...no, it is definitely not normal.
00:27:34.400 --> 00:27:38.600
So, it is not normal; oh, no, what are we going to do?
00:27:38.600 --> 00:27:43.100
Well, the next thing we ask ourselves is, "Can we cancel factors?"
00:27:43.100 --> 00:27:52.000
Can we cancel? Not easily: 1/(x + 3) - 1/3...I really don't see any easy ways to make cancellation show up there.
00:27:52.000 --> 00:27:55.400
So, we are probably not going to be able to cancel, at least not easily.
00:27:55.400 --> 00:27:58.500
So, our next question is...last time we asked ourselves, "Can we rationalize?"
00:27:58.500 --> 00:28:01.000
Well, there are no radicals here, so we can't rationalize.
00:28:01.000 --> 00:28:03.800
But we can take a hint from the idea of rationalization.
00:28:03.800 --> 00:28:09.000
The idea of rationalization was to multiply the top and the bottom by something that makes some part not weird anymore,
00:28:09.000 --> 00:28:13.400
not as strange to deal with, so that hopefully, we can get cancellation to appear later.
00:28:13.400 --> 00:28:19.300
Well, what would make the top...the thing that is really strange about this is that we have fraction over fraction.
00:28:19.300 --> 00:28:26.000
We don't like fractions and fractions; so how can we get rid of some of those fractions by multiplying?
00:28:26.000 --> 00:28:31.700
Well, the easiest way to get rid of the denominator in the top is to just multiply by the denominators in the top.
00:28:31.700 --> 00:28:42.800
If we multiply...I will rewrite the thing out...limit as x goes to 0 of 1/(x + 3) - 1/3, all over x:
00:28:42.800 --> 00:28:49.800
well, what would get rid of the x + 3? x + 3 would get rid of the denominator of x + 3.
00:28:49.800 --> 00:28:53.300
What would get rid of 1/3? Well, multiply by 3.
00:28:53.300 --> 00:28:57.900
We can get rid of both of those denominators by multiplying that whole top by (x + 3) times 3.
00:28:57.900 --> 00:29:00.300
That will cancel out each of the denominators as we work through it.
00:29:00.300 --> 00:29:07.400
And remember: it is always going to multiply the whole thing; when we multiply, we multiply the quantity, because we have to have distribution showing up.
00:29:07.400 --> 00:29:14.300
And on the bottom, we will have to have the same thing, because otherwise we are changing the expression; we can't change the expression.
00:29:14.300 --> 00:29:20.300
x + 3, times 3, over x + 3, times 3; great; our limit continues...limit as x goes to 0...
00:29:20.300 --> 00:29:26.900
What do we get on the top? Well, (x + 3) times 1/(x + 3)...those cancel out, and we are left with just the 3 left over.
00:29:26.900 --> 00:29:33.700
So, (x + 3) times 3, on 1/(x + 3)...the (x + 3)s cancel out; we are left with just 3.
00:29:33.700 --> 00:29:45.300
Minus...when (x + 3) times 3 hits 1/3, well, the (x + 3) doesn't do anything; but the times 3 cancels out, so we are left with minus (x + 3).
00:29:45.300 --> 00:29:49.400
All right, now we could expand the bottom, but that won't actually help us.
00:29:49.400 --> 00:29:53.500
One of the ideas that we are going to hopefully manage to get to is to figure out a way to cancel things.
00:29:53.500 --> 00:29:59.300
We couldn't cancel things easily by factoring; but hopefully we will still manage to cancel something at some point later on.
00:29:59.300 --> 00:30:04.900
We don't want to expand factors; we want to actually keep up this process of keeping things in factors.
00:30:04.900 --> 00:30:10.100
So, let's not put anything together; we will have it as x times x + 3 times 3.
00:30:10.100 --> 00:30:14.600
At this point, we see x + 3 on top and x + 3 on the bottom, but we have to be careful; don't cancel stuff.
00:30:14.600 --> 00:30:18.400
We can't cancel, because there is still a subtraction sign on the top; we have to have the whole factor.
00:30:18.400 --> 00:30:25.200
We keep working to simplify: limit as x goes to 0...3 - (x + 3)...well, the 3's will cancel out,
00:30:25.200 --> 00:30:37.800
and we will be left with just -x on the top, divided by x times (x + 3) times 3; great.
00:30:37.800 --> 00:30:43.700
Limit as x goes to 0 of -x/x(x + 3)(3); at this point, we say, "We can cancel some stuff!"
00:30:43.700 --> 00:30:52.100
This x and this x cancel, and we are left with the limit as x goes to 0 of -1 now (because it just canceled out the x,
00:30:52.100 --> 00:30:56.400
not also the negative), times (x + 3), times 3.
00:30:56.400 --> 00:31:01.800
Now, we ask ourselves, "Now that we have managed to cancel something, if we were to plug in a number, would we have something weird happen?"
00:31:01.800 --> 00:31:09.400
Would it be normal now? If we plug in 0, we get -1/(0 + 3)(3); it doesn't look like we are going to be having dividing-by-zero issues anymore.
00:31:09.400 --> 00:31:18.800
It isn't weird anymore, so we can just plug in; this is equal to the -1, over (0 + 3) times 3.
00:31:18.800 --> 00:31:22.500
Once it is not weird, we can plug in, because now it is effectively normal.
00:31:22.500 --> 00:31:27.300
And when it is effectively a normal function, you can just plug into it with your limit.
00:31:27.300 --> 00:31:36.500
-1/3(3) gets us -1/9; and there is our answer--great.
00:31:36.500 --> 00:31:40.200
All right, at this point, we have a really good understanding for how to figure out how limits work.
00:31:40.200 --> 00:31:44.800
The basic idea is, "All right, I have a limit that is normal and doesn't have anything weird happen."
00:31:44.800 --> 00:31:48.400
That is easy--just plug in something and crank it out--see what number you get out.
00:31:48.400 --> 00:31:52.000
That is what the limit is, because "normal" means that your expectations will be met.
00:31:52.000 --> 00:31:56.500
If it is not normal, if there is something weird happening, you try to manipulate things.
00:31:56.500 --> 00:32:02.500
You either pull out factors or multiply the top and the bottom; you do something where you are allowed to cancel factors later on.
00:32:02.500 --> 00:32:08.900
And then, you check and see, "OK, now that I have canceled out the factors, is it possible for me to plug things in and have it be normal, effectively?"
00:32:08.900 --> 00:32:12.600
Can I now plug in (now that there isn't, hopefully, a weird thing happening)?
00:32:12.600 --> 00:32:19.000
Sometimes there will still be weird things happening; in all of the examples we saw here, we canceled out anything that would cause weird stuff to happen.
00:32:19.000 --> 00:32:22.700
But sometimes, you will end up still having weird stuff, no matter what you manage to cancel out.
00:32:22.700 --> 00:32:27.300
And in that case, it can help to check a graph and think, "Oh, I see: it is going to go out to infinity," or something like that.
00:32:27.300 --> 00:32:29.100
And you will see that it is never going to work.
00:32:29.100 --> 00:32:34.100
But a lot of the time, you can cancel stuff out, and you can say, "Oh, OK, now it is effectively behaving like a normal function;
00:32:34.100 --> 00:32:37.900
so I can plug in the x-value that I am going towards and just crank out an answer."
00:32:37.900 --> 00:32:40.000
All right, we will see you at Educator.com later--goodbye!