StudyForge Samples > Chapter 11: Calculus Samples > Lesson 1: Introduction to Limits

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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 probably 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 the slope, at a certain point?

Now up to now, the only way we can do that is to estimate the tangent line,

So we just kind of approximate with that 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:

The slope of that secant line is just rise over run.

Or delta y over delta x.

Which of course is y 2 minus y 1 over x 2 minus x 1.

Which in this case is a squared minus 1 over a minus 1.

And that's the equation for our slope of the secant line.

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 squared minus 1 divided by 1.5 minus 1 and we'd actually get a slope of 2.5

And that would be our approximation

Well let's make it even closer to 1

How about 1.3?

Well if a was 1.3 and we plugged it into our slope equation

then we’d actually get a slope of 2.3.

And if a was 1.2? Then the slope would actually turn out to be 2.2

And if a was even closer, say 1.1, the slope actually is 2.1

And you can kinda see where this is going

So what if we picked something like 1.005? Like really close to 1.

Well when you calculate your slope it's 2.005.

And if you kept doing this you can already see that the slope seems to be getting

closer and closer to a LIMIT of 2.

But the problem is, we can’t just plug in “1”,

wish we could do that but if we did that we'd actually get a zero in the denominator

and we'd have an error.

But wait! Look at this, we can actually factor the numerator.

That a squared minus 1 is a difference of squares.

It factors into a minus 1 a plus 1.

And then of course the a minus 1 divides and we're left with a plus 1.

Now, we still can’t plug in 1 for “a”,

cause it just came from an equation where you can't have that, but that’s ok,

because we're not actually making a equal one we are just getting INFINITELY CLOSE to “1”.

And so we could say something like this:

The slope of our tangent line is gunna equal the slope of that secant line

when a gets infinitely close to 1.

Or in other words, the slope of the tangent line is going to equal a plus 1

when a gets infinitely close to one.

Well when a gets infinitely close to 1,

you're going to basically have 1 plus 1.

Or if you like 1.00000 etc plus 1 which is gunna equal 2.

And there it is, the slope of our tangent line equals 2.

It's a little bit shady there, but we'll do it more tight mathematically,

but that's essentially it.

Now, you might be like, ok, that was just lucky though

because we could actually factor the numerator.

Honestly, it 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.

So it's not kinda of so shady. So here's actually how it works.

Do you see where we have

the slope of the tangent line is equal to the slope of the secant line

when a gets infinitely close to 1

well, we’d actually write that like this:

The slope of the tangent line equals the limit of the slope of the secant line

as a goes to 1.

Then on the next line we'd actually have the limit of a plus 1 as a goes to 1.

And then of course the next line we'd just go equals 1 plus 1 equals 2.

And that would be proper mathematically.

Now lets have a look the following diagram.

In the diagram, you can see we have a hole as x=a.

Now we can't say anything about the actual y value at 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 gets infinity close 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 there

and in this case that it actually is no y value at all, nope, 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 rigouress 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 f of x as x approaches a is L

or another way of saying the thing is f of x goes to L

or the y value goes to L as x goes to a.

So, for example 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 closer to 1,

from either side, as it get closer and closer and closer to 1.

It gets closer and closer, closer to a y value of 2

and that means of the limit actually equal 2.

Now if we look at the 3, as x get closer and closer and closer to 3.

You can see that the y value is getting closer and closer and closer to 4.

So, the limit is 4. Now, you might be like,

but wait a second, they actually no value there,

at x equal 3, but with limits, which is part of the strength of the limit,

is we actually don’t need that point to exist we just get infinity close,

and so the limit is still 4.

Okay, what about this example

Let’s do the limits as x goes to 0, 1 and 2 all the way up to 5 for this function,

and I purposely made this one really screwed up.

So, you can see how kind of limits basically work.

Ah, let’s start with x goes to 1 now that one is kind of similar to the example we just did,

you can see as x gets closer and closer and closer to 1 from either side,

it gets closer and closer and closer to a y value of 3.

So, actually the limit of x goes to 1, is 3.

Now you also know that the actual point there,

the actual y value is actually 2, when x is 1,

but we don’t care right, because we don't care actually what is happening at that point,

we just care what happens at you get infinitely close to that point.

Okay, let’s move on, let go to the limit as x goes to 2

and you can see from the left or the right as you get closer and closer and closer to 2,

you get closer and closer to y value of 4, in fact you're already at y value of 4.

So, there you go, yeah.

On the next one,

the limit as x goes to 3 from the left, from the left you're getting closer and closer to y of 4,

and you already at 4 entire time.

But from the right, you actually getting closer and closer and closer to y value of 2.

Now the way you get write that is like this,

of a limit of f of x as x goes to 3 from the left,

pull up negative sign there, equals 4.

Limit of f(x) is x goes to 3 from the right, a plus sign, equals to 2.

But because those 2 limits are not the same,

4 from the left, and 2 from the right,

that means our limit actually does not exist.

So, no limit there.

As x goes to 4, we actually see that from the left,

these thing goes down, down, down to negative infinity,

and from the right, as it get’s closer to 4 it also looks down to negative infinity.

And some people might it say that infinitely technically is not a number and therefore doesn't exist,

Ah, I think it just nice to actually write down that our limit as x goes to 4 is simply negative infinity.

That’s because the it's the same from the both sides.

Right, it was different from both sides like the previous one, you'd say that the limit DNE.

Now for x goes to 5 . Umm you kinda only have a limit from the left hand side right?

All you've got is the limit as x goes to 5 from the left,

and that's of course approaching y value 1.

And so you if only have one side that's just fine and you'd say your limit actually equald 1.

Okay, of course going back to the final one at x goes to zero same deal,

but this time we only have a 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 limits.

So, here they are:

Suppose we move x infinitely close to value of a from the left.

Then, if the y value getting infinitely close to L as we do that, then we'd have this:

The limit is x goes to a from the left of f(x) equals 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 handed limit,

and for the limit to exist at a certain point,

both the left sided limit and the right sided limit must be equal to each other,

or in other words, if the limit from the left equals L

and the limit from the right equals L,

then we can say that the limit as x goes to a is L,

of course that’s providing each of those one sided limits actually exists,

for instance in the last example when you looking at x equals 5

all we have the left handed limit, but that okay right? Like,

it was the end point and there wasn’t the right hand limit and that’s totally fine.

Okay, in this lesson, we're gunna do a whack of limit questions!! So here we go!

So for our first question: a little trick.

If you can, just stick in the value for x and if there is no problems,

like no square roots of negatives or no dividing by zeros,

hey you’re done.

So when you stick in 2 here in this question you get,

2 squared plus four times 2 minus 3 which is 4 plus 8 minus 3 which is 9,

no problems, there you go. There is your limit.

And if you wanted you could actually graph y equals x squared plus 4x minus 3,

and then you’ll see as x gets close to 2 you’ll get close to a y value of 9.

Alright next question if you stick in pi over 2 for x, do we get any problems?

No we do not, so that’s exactly what we are going to do.

The answer for this one is gunna be the sine of pie of 2 divided by pie of 2,

and that's going to be 1 over pie of 2 which is 2 over pie. Done.

Alright next one, so for the limit of x goes to 2 we kinda have two different sides to consider here,

we have the left hand side limit and we have the right side limit.

For the left hand side limit, you can see that it’s shooting down to negative infinity.

For the right sided limit, it’s actually going to positive infinity.

Those two numbers are different, so this limit does not exists.

Now for x goes to 3, again from the left hand side it’s approaching a y value of 1,

but on the right hand side, it’s approaching a of y value 2.

So those one sided limits, the left one and the right one are different,

so again, this limit does not exist.

Alright next question, so again we try to always stick in the value for x if we can

as long as there’s no problems.

Now here when you stick in 0 for x, we are dividing by 0 and that’s a problem.

So we do the right and the left handed side.

Ok so first we’ll do the left sided limit, so we’ll do the limit as x goes to 0 from the left.

It’ll be the absolute value of x over x.

Now in the bottom you will get some number like close to 0 but its negative right?

Maybe like negative 0.0001 and on top you’re taking the absolute value of that,

so it’s actually gunna be positive.

So you’re going to get the same number but a positive on top and negative on the bottom

and when those divide you’re actually going to get negative 1.

So the way I kind of write it is like you’ll get negative 0 on the bottom and positive 0 on top

and you when you divide that’s negative 1.

But of course you shouldn't really write positive 0 over negative 0

but anyway mentally that’s what you’re doing.

Then let’s do the right sided limit,

we have the limit of x goes to zero from the right hand side and here again we stick in some number close to 0

say .00001 but its positive each time

and that absolute value isn't going to make any difference

so we actually get like positive 0 over positive 0 which is going to just give us 1,

but again don't do the positive stuff.

Then if you look it at that the left sided limit and right sided limit are different and so this limit does not exist.

For the next question, we have the limit as x goes to 0 of sine x over x.

Now again you try to stick in 0 for x but you get a 0 in the denominator so you can't do that.

And then you do your left sided limit and right sided limit

and for this when there is kind of no other way to do it other than to just get a calculator out

and plug in some number really close to 0 from the left so let’s do negative 0.00001

and when you just stick that in, into sine x over x you could get number really really close to 1.

So your limit is actually going to equal 1.

For the right sided limit,

again same we’ll stick a number in this time positive and so let’s try .00001

and we stick that in and again same thing you actually get a limit of 1

and so both the left and right side limits of the same, as so this limit equals 1.

Alright, let’s keep going.

So next question we have tan x over x again you can't just stick in 0

and if you want you can do previous trick where you just plugin numbers that is really really close to 0

but I want to show you an little algebraic trick here that you can do,

kind of fun, this is kind of dipping of little bit into the next lesson but that's ok,

we’ll explain this a little more formally then.

But tan x as you know is sine x over cos x

so we can actually break this up a little bit and write that as sin x over cos x times 1 over x.

Now we can rewrite that little bit, that's the same as sin x over x times 1 over cos x

and the rules of limits are if 2 limits exist you can actually break them out in the separates limits.

So this actually is going to equal

the limit as x goes to 0 of sin x over x times the limit as x goes to 0 of one over cos x

because both of those individual limits exist.

The first one is limit we just did.

We know that actually equals 1.

You should actually commit that to memory, that is a common one.

The limit as goes to 0 of sin x over x equals 1.

In the next one, we can actually just plug in 0 into x, and when you have cos 0, which is just 1,

this overall limit is going to equal 1 times 1 which is one and there is your limit.

Ok, so this question of the limit as x goes to 0 of 3 over x squared,

you can't just plug in 0 because you’ll have a problem in the denominator.

But let’s be honest here, when you take 0 from the right or the left and you square it,

it’s going to be positive.

So, what we end up with here is with 3 divided by a very positive small number.

And so the way I say it is when anything divided by small is big.

Anything divided by very small is very big.

And so here we have 3 divided by 0 which is basically going to be infinity or positive infinity and that's your limit.

Ah the next question,

the limit as x goes to 0 of 4 over x and you kind of want to do the same thing that you just previously did

but the problem is if you go from 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, right?

because you going to have 4 divided by negative 0 essentially

and 4 divided by negative 0 would give you negative infinity

and from the right hand side, you get 4 divided by positive 0,

that's give you positive infinity and those two limits are not the same,

so that limit does not exists.

- http://tube.geogebra.org/student/m118774 (Press Play to see the fun!)

- An introduction to limits from an intuitive point of view

http://archives.math.utk.edu/visual.calculus/1/limits.16/index.html

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 resources 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 special 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.)