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So, in order to delve more into the Calculus
(yup, that’s right, it was originally called “The Calculus” meaning “The Calculator”!),
we need to be able to zoom in infinitely close to a certain point to see what’s really going on there,
and add up infinitely many little areas to get the whole area,
or, well, just tons of infinite stuff, really.
And so before we can really get into the Calculus,
we really need to understand the whole concept of infinity and everything about that.
Now there was this ancient paradox called Zeno’s Race Horse Paradox, dating around 450 B.C.
Basically, it went like this: You’ve got a racehorse in a 100m race.
Now the racehorse first covers half the distance to the finish line (which is 50 m) in say, 4 seconds.
And then it would cover the next half (25 m), in about 2 seconds.
And then it would cover the next half, 12.5 metres, then the 6.25 metres,
then 3.125 metres, and so and so on, each time covering half of the remaining distance,
and this way, never, ever actually getting to the finish line. ??
And this is a paradox, of course,
because everyone had seen the races and knew the horse always DID cross the finish line!
And yet if you looked at it with this halving idea,
the horse never seems to get there.
And, they never did solve the paradox,
and hence why this is the end of your Calculus course. Congratulations.
Okay, they DID actually eventually get around the paradox,
but it took them hundreds of years to do it.
And what they did essentially, is come up with the idea of the limit.
That is, if you add up all of the distances, to infinity (as in, ALL of them) then eventually,
it adds up to a finite number, 100 m.
So the 50 + 25 + 12.5 + etc,etc,
all they up to infinity with add up to 100.
And similarly, if you did this with the time it took
and added those up to infinity (4 + 2+ 1+ 0.5 +0.25 + ...) etc.
they would add up to a finite number as well, 8 seconds.
So in this chapter, we’re going to start with limits and then get right into the Calculus.
Set your pacemakers, ‘cause here we go.
In this lesson, we’re going to explore the basic idea of a what a limit is.
To start off, have a look at this curve and drag the point around a bit,
and as you do, notice the tangent line at that point.
Notice how the tangent line changes steepness depending on where the point is?
Oh, I should give you the definition of a tangent line.
A tangent line at a certain point is a line that best approximates the steepness of the curve at that point.
So here’s a simple question:
What’s the exact steepness, or slope, at a certain point?
Now up to now, the only way we can do that is to estimate the tangent line,
So we can approximate with the tangent line will be,
and then do a little slope triangle thing rise over run,
and calculate our approximate slope.
But if we’re talking about the orbit of a satellite in outer space,
or the strength of a bridge under the stress of a semi-truck,
approximations will get people killed!
And we don't want a killed anybody here, in this course.
So we need something better. Enter, the limit.
So for Ex. For the curve y = x2, what is the value of the slope when x is 1?
Okay, here’s how we’re going to do this.
We’re actually going to start with an approximation of the tangent line by picking another point close to x = 1
and then we’ll connect them with a line, which will give us a secant line.
(Remember, a secant line is a line that connects two points on a curve.)
Now we have the secant line, which is an exactly tangent line but it is short of close.
Then we’ll calculate the slope and that’ll be our estimate for the actual slope of the tangent line at x = 1.
Of course, you can see that it’s a pretty poor estimate of the actual tangent line.
What we need to do is move the point closer, and then we’ll have a better estimate.
Actually, why don’t you play around with that other point
and move it closer to x = 1 to see how the secant line can approximate the tangent line
if you move it close enough.
When you’re ready to proceed, click to continue.
Ok, so here’s how we’re going to do this mathematically.
First, we’re going to go back to that other point we picked,
say it has an x-value of “a”,
and we’ll get its corresponding y-value by putting it into the equation for the curve, y = x^2,
and we get a y-value a^2. And so we have the two points (1,1) and (a, a^2).
Now, can you see what we’re going to do next?
We’re going to set up the slope equation using those two points,
and then, can you guess it?
the suspense is building, we’re going to make “a” get very, very close to 1,
and we do voila, we’ll have our slope of the tangent line!
If you didn’t catch that, here’s what I mean:
And now we do we just make “a” get closer and closer to an x-value of 1.
For example, what if a value was 1.5?
The our slope would be
1.5^2 – 1 / (1.5-1) = 2.5
What if a were 1.3? Then we’d get a slope of 2.3
What if a were 1.2? Then the slope would be 2.2
the slope would be 2.1
What if we kept going, and a was like, 1.005?
Then the slope would be 2.005.
Etc, etc, and you can see that the slope appears to be getting close to a LIMIT of 2.
But rats, we still can’t just plug in “1” for “a” because we’d get an error,
with a zero in the denominator.
But wait! Look at this, we can factor the top!
Now, we still can’t plug in 1 for “a”, but that’s ok,
because we’re simply getting INFINITELY CLOSE to “1”, not actually getting there.
And so we could say something like this:
(thought it doesn’t actually get there)
Wow, that was satisfying.
Now, you might be like, ok, that was just lucky that we could factor the numerator there.
Nope, turns out you can always do some kind of math thing,
usually factoring, to make this kind of thing work,
and then you just plug in your value for a, and you’re done!
Now, just before we finish this lesson,
I would like to show you how we say that limit thing properly. See where we have
mtan = msec when a gets infinitely close to 1
well, we’d actually write that like this:
mtan = lim msec (“The limit of the slope of the secant line, as a goes to 1” – for me to say)
= lim (a + 1)
Now we have been to introduce of another idea of limits .
Its very important that we understand
how it works mathematically is so help up
solve actually problem is about to face has worst cases ready for the calculus.
Now lets have a look the following diagram.
In the diagram, you can see we have a whole as x=a.
Now we can't say anything about the actual y value with that point since it doesn't exist.
What about the y value really close to that point,
that is what's the y value when x get s infinity closed a.
If it get actually gets closer and closer and closer to specific y value,
then this y value is called the Limit.
Do we actually care that it never gets their
and in this case that it actually is no y value here at all, no, not at all.
The limit doesn't care about the actual y value at a point.
This is really important let me say that again,
the limit doesn't care about the actual y value at a point.
Its only care about what happens as it you get infinitely close to a point.
Now here is regress definition of the Limit, which is fortunately pretty similar.
Suppose you move x infinitely close to a, but not equal to a,
if the y value to get infinitely close to l,
as we do that and we say a limit of above x is approaches a is L
or another way of saying the thing is above x goes to L
or the y value goes to L as x goes to a.
So, for e.g. in the following graph,
what the limit of f(x) as x goes to 1,
and what the limit of f(x) as x goes to 3,
well looking at x goes to 1, you can see that x get close to 1,
from either side and I think it’s closer, closer and closer to 1.
It get closer and closer, closer to a y value of 2
and that means of the limit actually equal 2.
Now for looking the 3, x get closer, closer, closer 3.
You can see that the y value is getting closer and closer, closer to 4.
So, the limit is 4, you might be like,
but wait a second, they actually no value there,
as x equal 3, but with limit which is part of the strength of the limit.
So, actually we don’t need that point to exist we just get the infinity close,
and so the limit is still 4.
Okay, what about this e.g.
Let’s do the limit is 2 goes to 0,1,2,….5 for this function,
and I purposely made this, really scrod up.
So, you can see how kind of now limits basically works.
Ah, let’s start with x goes to 1 now that one is kind of similar e.g. we just did,
you can see as x get closer, closer, closer to 1 from either side,
it get closer, closer and closer to y value of 3.
So, actual the limit of x goes to 1 is 3.
Now you also know they actual points there,
they actual y value is actually 2.
When x is 1, but we don’t care what actually happening at that point.
We just care what happens at you get infinitely close to that point.
Okay, let’s move on, let go the limit as x goes to 2
and you can see from the left or the right as it closer, closer, closer to 2.
You can get closer, closer to y value of 4, innately at y value of 4.
So, there we go, yeah on the next one,
the limit as x goes to 3 from the left, from the left you can get closer, closer to y of 4,
and you already at 4 entire time.
But from the right, you actually getting closer, closer, closer to y value of 2 way you get right that.
Is like this, of the limit above x as x goes to 3 from the left,
pull up negative sign there, equal 4.
Limit of f(x) is x goes to 3 from the right plus sign equal to 2.
But because those 2 limit are not the same,
4 from the left, and 2 from the right.
That means the limit actually DNE.
So, no limit there, as x goes to 4, we actually see that from the left,
these thing goes down, down, down to negative infinitely,
and from the right, as it get’s closer to 4.
It also looks down to negative infinity.
Some people might it say that infinitely technically is not the no, so DNE,
Ah, I think it just nice actually right down that our limit is x goes to 4 is simply negative infinity.
That’s because the same from the both side,
right it was different from both side that the previous one,
intensively limit DNE.
Now for x goes to 5 . Umm you can only have a limit from the left hand side right.
All you got is the limit as x goes to 5 from the left,
and that of course approaching y value 1.
And so you only have one side that just find and you say your limit actually equal 1.
Okay, of course going back to the final one at x goes to zero same deal,
but this time we only have right hand limit,
So, x goes to zero from right, you can see is also getting closer to y value of 1.
And so we say the limit is actually one.
Now, we should do a formula definition for those one sided limit.
So, here they are:
Suppose we move x infinitely close to value of a form the left (but not equal to a).
Then, if the y value getting infinitely close to L as we do that, we’d say:
The limit is x goes to a from the left of f(x) is equal to L
and same is the opposite right if we move x infinitely close to a from the right
and y value is getting closer, closer to L.
Then there you go, you got the right hand of limit,
and for the limit to exist at a certain point,
both the left side limit and the right side limit must be equal to each other,
or in other words, if the limit from the left is equal L
and the limit from the right is equal L,
then we can say that the limit as x goes to a f(x) is L,
of course that’s providing each of those one sided limits actually exists,
for instead in the last e.g. when you looking at x equal to 5
all we have the left hand limit, but that okay right.
It was the end point and there wasn’t the right hand limit and that’s totally fine.
Okay, in this lesson, we’re gonna do a whack of limit questions!!
Here we go!
So first question a little trick.
If you can just stick in the value tracks it is no problems like no square to negative
or no dividing by zeros,
hey you have done.
So stick in 2 here in this question you get 2 squared that's four times 2 -3
which is 4+8-3 which is 9 no problems here are you go.
There is your limit.
That you want you crushing crap y=x creeper 4x-3
then you see x=2 it get close to while thy of 9.
Ok next question if you stick in pie 2 for x,
do we get any problem, no we do not, so that exact what we gonna do.
The answer of first one is gonna be a sign of pie of 2 divided by pie of 2
that's gonna be 1 for pie of 2 which is 2 over pie.
Done. Alright next one,
so for the limit of x goes to 2 we can't have two different sides to consider here
we have left hand side limit and we have the right side limit.
For the left hand side limit,
you can see it should down to negative infinity.
For the right side limit, that is going to positive infinity.
Those two numbers is different, so this limit does not exists.
Now rest goes to 3,
again from the left hand side is a pursing of y value 1
but on the right hand side, is a pursing of y value 2.
So those one sided limits left on the ray one different so again this limit does not exist.
Alright next question,
so again we try to always taken of the value frac if we can form is no problem.
Now here is taken 0 for x we dividing by 0 in a problem.
So we go the right and left hand side.
Ok so first with little left sided limit,
so the limit is x=0 through left we have yx over x.
Now in the bottom you will get some number like =0
but its negative write may be like negative 0.0001
and on top taken the outside value that so I,
share gonna be positive to gonna be same number
but a positive is on top and negative from the bottom those divide here sharing get negative 1.
So we x had a right is like you get negative 0 in the bottom
and positive 0 on top and you define that negative1.
But of course you shouldn't very very confidence 0 is a negative 0
but anyway mentuless doing.
Then is to the right side of limit,
limit is x=0 right hand side
and here again we taken sub number for close to 0 say .00001
but its positive reach time
and that absolute value isn't gonna make any different
so splash give positive 0 over positive 0 which is gonna give us 1 well don't do the positive side.
Then if you look it that a left side a limit
and right of limit a different and so this limit does not exist.
Next question, we have the limit x=0 of sign x over x now again you try to stick in 0 for x
or she gets a 0 of ramrets it can't do that let it do your left sided limit
and right sided limit
and through this for the kind over the way to do it other then sticker a calculator out
and plug and some number really closed to 0 from the left so lets do negative 0.00001
and you to stick that in the side x over x you could get number really really closed to 1,
so your limit is actually get equal1.
For the right sided limit, again same thing to stick a number in this time positive
and so try .00001 and stick that end
and again same thing you get actually get a limit of 1
and so both the left and right side limits of the same into this limit equals to 1.
Alright, just keep going.
So next question we have 10x over x again we can't just taken 0
the more you can do previous trick were you just plugin numbers
that is really really closed to 0
well I am not sure you really out of break trick here
that you can do a kind of find there was kind a dipping of little bit into that x lesson
but that's, ok looks play in this little bit more performer you can.
But 10x as you know is sign x over cos x so we can actually break this up a little bit
and write their as sin x over cos x tan over x.
Now we can rewrite a little bit that's the same as sin x over x times 1 over cos x
and the rules of limits kind of 2 limits exists
you could actually break them out in the separates limits.
So this actually is gonna equal to limits as x=0 of sin x of braces times limits is x=0 a
one over cos x because both of those individual limits exists.
The first one is limits we just did.
We know that actually equal's 1,
usally committed tumarious is the common one.
Limit x=0 sin x over x =1.
In the next one, we can actually just plugin 0 and x,
we have of course 0 which is just 1 this overall limit is gonna equal 1 tan
which is one and there is your left.
Ok, so this question of limits is x=0, 3 over x square we can't just put it 0
because your problem is nominator.
Well lets be honest here,
when you take 0 on right or the left
and you square it, its gonna be positive.
So, we end up here is with 3 divided by a very positive small number.
So when I says anything divided by small is big.
Eighteen divided by very small is very big.
Here we have 3 divided by 0 which is basically gonna be infinity
or positive and that's your limit.
A next question, x=0 4 over x
and we can't wanted to same as just previously did the problem
and if you go for the right or the left,
you not gonna get the same positive thing here.
From the left hand side,
you actually get a negative number write
because you can a have 4 divided by negative 0 essentially
and the 4 divided by negative 0 would be negative infinity
and from the right hand side,
and from the right side 4 divided by positive 0,
that's give you positive infinity ans those two limits are not the same,
so that limits does not exists.
The above paradox was originally proposed by Zeno of Alea in a more general form, in the sense that continually halfing a distance will result in never achieving the whole. It has been expressed through various means, including races such as the above, and was also used to explain the inability of the warrior Achilles to catch a turtle: if the warrior advances even slightly, so will the turtle, and thus he will never catch up. "And thus in every time in which what is pursuing will traverse the [interval] which what is fleeing, being slower, has already advanced, what is fleeing will also advance some amount." - Simplicious, from "On Aristotle's Physics"
Some resources/professors will say that for an endpoint of the domain (such as x=0, 5 in the Ex. on Page 3), that because you don't have both a left and a right-hand limit, that the limit doesn't exist; other resoures or professors will say that it does exist, and is simply equal to the one sided limit there. The problem is that either way, it's actually a point of mathematical inconsistency in Calculus, since a speciall allowance for endpoints will have to made at some point, based on current definitions. Those who adopt the former definition have to make these special allowances for endpoints when it comes to the upcoming topic of Continuity, not to mention Differentiability afterwards, whereas those who adopt the latter convention, as we have, make the special allowance only at Limits, and then it's consistent from that point on, not to mention more intuitive. (We won't ask any questions on that specific idea here though, to respect the differences.)