Of course, as we've seen,
the idea of calculation functions limit by definition,
it's tricky and takes a lot of time.
So people don't usually use the ideas that we need
to come up with the limit
and they're proved by definition.
Here comes our basic arithmetic rules.
We've already seen them in case of sequence of limit.
So let us just to reiterate it, to revisit,
and cover it once and for all.
For example, basic rules here are the limits of
some arithmetic operation is
there is the arithmetic operation of these things.
This works for sum, difference,
products, and division of functions.
Well basically, the limit of
a sum is the sum of limits, and so on.
But here comes the tricky part.
Of course, you shall always
think about the division because division
actually produces a headache for us
because it crucially affects the mean and all the others,
well you can't divide by 0.
This is the school rule.
So the thing should be valid about that,
in the denominator, you should always get non-zero value.
So basically, when we say that as
the limit of division is the division of limits,
we should require that
the limit of the denominator in
the initial fraction doesn't approach 0.
That's a nice try.
Here is another thing, basically,
what we should expect that
the initial function, this way,
fraction f divided by g is
also defined in some neighborhood
of the limits point, basically,
it means that there is a neighborhood of
the point where function
g towards x has no zero values because otherwise,
we have in every neighborhood,
some crazy point with no value,
and we cannot proceed with
our epsilon-delta from our definition of the limit right.
So that's a common thing, we've already seen them,
and we are going to look at for example pattern,
but let's stress out one more thing,
and it's the most important one here.
All these rules are true only in
case all right hand side limit exists.
So all of these is correct only if these two terms exist,
basically, it means that if you
assume that you have some finite limit function,
for example, more function equals to one.
You can't change it into, for example,
x plus 1 minus x,
and can see the a,
for example, plus infinity,
and say that this holds.
Infinity is not a limit,
it's not a finite limit,
so it doesn't exist.
This is a crucial rule that you must remember.
So let us look for example some way to
see like the limits are fraction x squared minus 1,
and x power 3 minus 1.
So the limit points here is, well,
just for now, for example, is 0.
So what we should think firstly?
Let us try to calculate the limit by our arithmetic rule.
For example if x approaches 0,
then x approaches 0.
It's quite easy.
If x approaches 0,
then x-squared is basically x multiply by x.
Thus by our product rule,
it also approaches 0 multiplied by 0.
Same thing applies to x power 3,
and that's what we get.
We get basically that our nominator approaches 0 minus 1,
our denominator approaches 0 minus 1,
and the answer is by our arithmetic rules is 1.
That's quite nice.
Please know that all necessary conditions for
our division rule is actually satisfied
because we do not consider dividing by zero at all.
That's nice. But what if our limit point is trickier?
So we've seen this function with a limit of 4 and 0,
but now let us assume the limit when x approaches 1.
This is understandably much more verse,
because if x approaches 1,
then x-squared and x power 3 approaches 1,
and then our nominator and denominator approaches both 0,
and thus we get something which
previously was called indeterminate forms.
You remember 0 divided by 0 is our indeterminate form.
So what we are going to do in order to avoid this one,
and for the gap sake,
just get the answer for this example.
So then basically that
we are going to transform this function
with an equivalent function,
just by taking the factor
of both nominator and denominator.
So the nominator is basically multiplication of x plus 1,
and x minus 1,
and the denominator is x minus 1,
multiplied by x-squared plus x plus 1.
So if we divide
our nominator and denominator by their common multiply,
x minus 1, then we get a nice function x plus 1,
divided by x-squared plus x plus 1,
as x approaches 1,
the nominator approaches, let us see,2 I think.
1 plus 1, 2 and the denominator approaches 3.
Hence our answer is two-thirds.
That's fine. Let us take a look, for example,
at what we've done,
we've basically substituted our function,
our start,
our initial given function with some function here.
So are these functions the same or not?
Basically, you can understand that since
we've just divided denominator,
nominator by its common multiplier,
functions are pretty close, except one point,
and this point is x
equals to 1 because at this very point,
the initial function is not defined,
and the function we've
got during our transformation is actually well defined.
That has a value two-thirds,
which we have our answer.
So the thing here is that basically,
we've changed our function,
but we've changed our function only in one point,
one point is our limit point,
and this we've previously discussed.
We do not care about this variable limit points,
since our limit avoids
consideration of the valuer at the very limit point,
when we are talking about epsilon data formal definition.