One of them is talking about functional limits,
it's hard to not talk about two very basic since here.
Which are sometimes referred to as important or fabulous whatsoever.
There are two limits that everybody should know and
everybody does really sort of them at least quite a bit.
So the first one is already known for us is the definition of exponent, right?
Base of the natural logarithm whatsoever the E constant,
so what do we have in case of our sequences.
Now sequences with talk about that, if we're considering
the sequence 1 plus 1 divided by n powered n it approaches.
Some new, newly defined by
the resist relation constant e and that's kind of the definition, right?
So the trick there was that this thing is bounded and not only grows.
So thus, we have some kind of limits and
we should call it something if it is not a known number.
Okay, so let us make a simple transition into the case of function since
1 divided by n approaches 0 as n approaches infinity.
Then we can just substitute it with a real variable x.
Thus, we get 1 plus x power with 1 divided by x approaches
our e constant, if x approaches 0, all right?
So that's easy, that's quite nice,
that's basically generalization of what we already know.
The second limit that everybody knows and
there that we should talk about is a limit consultant in taken geometric rules.
Here basically the idea is that as
an angle approaches 0 at the sine of its angle is basically an angle itself.
Or in other words as the relation between sine of x and
x equals to 1 as x approaches 0, right.
So in order to get a sense get an understanding of what it is?
Let us do some basic picture about the areas on this graph.
For example, let us assume that
we are considering the circle with radius equals to 1.
So our unit circle and we are looking at the sector with an angle x, right?
So what's an idea here the idea here is that if x is measured in
radians then area of this vary sector is basically x divided by 2.
Since the area of the whole circle is pi as you do remember there was a circle is
pi multiplied by the right of squared, right?
So the area of this sector is proportionally half an x, right?
So we have two things here,
firstly we have a red triangle which lies within our blue sectors.
Thus, it has a small area and we have a green triangle visualize
which basically has a sector included in it.
Thus, it has a great area both sides of red
triangle equals to 1 and since we know the angle in between them.
Then the area of which is 1 multiplied by 1 multiplied by sine of its angle
half or just sine of x divided by 2.
That's nice, thus we get 1 by enter here and as for the green one,
we do know there's by definition one side of this triangle equals to 1.
So the triangle so far is square as of our circle and there are the sides is
basically by definition of tangent function is tangent of x.
Thus, in order to compute the area of this triangle,
we need to remember that it's orthogonal.
Plus we can just multiply these two sides and
divide them by 2 as we get a tangent of x divided by 2, right?
In other words is sine of x divided by 2 and
then divided by cosine of x.
Well since cosine function approaches.
Let us revisit it, you do remember as sine approaches 0 if
x approaches 0 sine cosine approaches 1 as sine approaches as x approaches 0, right.
Thus it approaches 0 and if we divide both sides by x for
example or by sine of x thus we get that sine of x divided by x approaches 1.
In the both sides, so we've actually just proven this very limit, that's nice.
So let us consider for example some example for it.
We are going to go for the combination of these two limits and
compute a limit as x approaches.
Infinity sine of 1 divided by x plus cosine of
1 divided by x in power of x, right?
So in order to do, firstly let us assume that
we just substitute 1 divided by x by a new variable t.
And t approaches 0 as x approaches infinity, right?
Thus, we have a nicer form sine x plus cosine x.
Powered 1 divided by t, just let us remind us that our new variable is called t.
All right, so in order to properly do it, we just let us revisit
the initial expression here and assume what we're looking at.
Just let us try to use our ultimate rules,
just can see the the limit of all the terms in here.
And to decide whether it is or not the indeterminate form, right?
So if x approaches infinity then sine 1 divided by x approaches 0, right?
Because sine of 0 is 0 same applies for the cosine,
cosine of 0 is 1 thus our break it approaches 1.
Thus, and all the function is basically the indeterminate form for
one part infinity, right?
So in order to avoid this we're going to to artificially yet
the form of our definition of e of our first important thing here, right?
So in order to do this, we need to artificially add and then subtract.
The number one in the bracket, right?
So we got 1 plus something, something
is sine t plus cosine t minus 1, right?
That's our kind of new x here, the x from there
in the power of 1 divided by t, right.
We know that if we divided it by the very
idea of 1 divided by x our new x, right?
That's we get an exponent here, so we need to do the same.
Basically we need to artificially form 1 divided by new x here,
in order to do that, we are going to cross the right exponent.
And then we're going to multiply our power here by sine t,
cosine t, minus 1 that's 1 this multiplier will get us an exponent.
And the rest of it will form sine t, plus cosine t,
minus 1 divided by t.
So is it enough?
Maybe we can just write the limit straight out loud,
well, I still kind of not the case for us.
Because let us look at it closely sine t
approaches 0, cosine t approaches 1.
Thus all the nominator approaches 0 and
t approaches 0 as we got in the terminal formed 0 divided by 0, right?
But if we just consider cosine t minus 1.
We can remember that this is basically the idea when
we have basic to be symmetric formula here, right?
Because 1 minus cosine is 2 sine
squared of 1/2 an angle, right?
And if we look at 2 sine squared divided by t.
It basically results into sine of half a t,
divided by half a t, multiplied by sine of half a t, right?
And if it is true thus, the first thin according
to our second important limit approaches 1.
The second sine approaches 0 thus the system
approaches 0 and the first term approaches 1.
Thus, we get exponent of 1 or just our constant e, okay.
That was rather complicated of us,
as you can see just substantial views of two important limits.
Can get us basically to the answer without any generalizations or [INAUDIBLE].
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