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.