Hello friends. So, welcome to lecture series
on multivariable calculus in the last lecture we have seen that what do we mean by a function
of several variables, domain and range of two or more than two variable functions. Whether,
it is open, closed, bounded, un-bounded. All these things we have seen in the last lecture.
Now, how can you find limit of multivariable functions. So, we recall it first you recall for limits
for one variable functions so, suppose we have limit x tending to a, f( x) is equal
to L. So, what does that mean it means that as x
approaches to a, f(x) approaches to L so, roughly speaking when we say that limit x
extend to a, f(x) equals to L. This means as f tending to a, f(x) tends to L, now how
can we defined this mathematically. So, the mathematical definition of limit is for every
epsilon greater than 0, there exists a corresponding real number delta greater than 0, such that
mod f(x) minus L less than epsilon whenever 0 less than mod x minus a less than delta.
You take any epsilon greater than 0, no matter how small how large epsilon you are taking.
There will always exist a corresponding delta greater than 0 such that f(x) minus L, the
mod of this less than epsilon whenever this quantity less than delta greater than 0 or
we can say that this implies mod of f(x) minus L less than delta. Now what does this definition
mean, now let us see. Now here we are having that limit x tending
to a, f(x) is equals to L, this means for every epsilon greater than 0 there exist corresponding
delta greater than 0 such that mod f(x) minus L less than epsilon whenever 0 less than mod
x minus a there is a delta. Now, this means that x minus a is less than
delta, greater than minus delta and this means that x is less than a plus delta and greater
than a minus delta, and of course, x should not equal to a because if x equal to a then
it is equal to 0 and so x not equal to a. This sometimes we call as deleted neighbourhood
this we are also called as a deleted neighbourhood of x at a.
Now, same similarly what this means what this inequality represent this is f(x) minus L
less than epsilon greater than minus epsilon or f(x) is less than L plus epsilon and greater
than L minus epsilon. So, this means now this is, suppose L, this is f(x) equal to L now
this is suppose L minus epsilon and this is suppose L plus epsilon. Now x is between a
minus delta to a plus delta, you take suppose this is x equal to a suppose this is a minus
delta and this is suppose a plus delta. Now, this definition means you take any epsilon
greater than 0, no matter how small how large epsilon you are taking there will always exist
a delta such that the image of all those x lying in this interval
will always be contained in this band means you take any epsilon greater than 0
no matter how small, how large you are taking, there will always exist a corresponding delta
such that image of all those x lying in this interval will always be contained in this
band. Now, if you take epsilon very small tending
to L tending to 0 then this delta will definitely tending to zero; that means, that as x approaches
to a f(x) will approach to L because as epsilon tend 0, this will tends to L and as delta
tend to 0, this will tends to a. So, x extend to a f(x) will tends to L so, we can say that
f of all those a minus delta to a plus delta this all those x lying in this interval will
be contained in L minus epsilon to L plus epsilon so; that means, for every epsilon
greater than 0 there will exist a deleted neighbourhood of x at a, such that the image
of all those x lying in this interval will be contained totally inside this band, totally
inside this interval ok so, this is how we can defined limit of single variable functions. Now, let us first discuss few examples based
on this for example, limit extending to 2, 2 x plus 4 plus 1 is equals to 5. This is
very simple you one can easily see that limit of this simple function 2 x plus 1, is x into
2 is 5 how can we prove this using delta epsilon definition.
So, let us see how can we prove so, let epsilon greater than 0 be given you take any epsilon
greater than 0 you take some epsilon greater than 0. The same process will repeat for any
greater than zero so, we can say that for epsilon greater than 0 there will exist some
delta. So, basically what we have to show, we have to show that for you take any epsilon
greater than 0 there will always exist corresponding delta such that definition holds. What we
have to prove basically that for every epsilon greater than 0 there will exist corresponding
delta greater than 0 such that mod of 2 x plus 1 minus 5, this is f(x) minus L less
than epsilon yeah less than epsilon whenever 0 less than mod x minus 1 less than delta.
So, basically x minus 2 less than delta so, basically we have to show the correspondence
of delta in terms of epsilon. We have to show the existence of delta so, how can we do that.
Now, we start with this inequality this is mod 2 x plus 1 minus 5, this is equals to
mod 2 x minus 4, this is equals to mod 2 x minus 2, this is equals to 2 mod x minus 2
and this is equals to if you take if this is less than delta. So, this is less than
delta. Now this quantity; this is less than 2 delta because this quantity is less than
delta. Now, if we choose 2 delta less than equals to epsilon, then mod of 2 x plus 1
minus 5 less than epsilon whenever 0 less than mod x minus 2 less than delta.
Now, this is because wherever you choose whenever you choose delta less than equal to epsilon
by 2 then this inequality always holds because this is less than 2 delta which is less than
equal to epsilon so; that means, this is less than this is less than epsilon whenever this
is less than delta. So, this inequality always holds whenever delta is less than equal to
epsilon by 2 you choose any delta greater than 0 you can always you can choose any epsilon
greater than 0, you can always find delta which is less than equal to epsilon by 2 for
which this inequality holds. So, we have shown the existence of delta in
terms of epsilon for different delta for different epsilon delta will be different, but we have
shown the existence of delta in terms of epsilon such that this inequality holds. Hence, we
can show we can say that this limit exists and is equal to 5.
So, this is how using delta epsilon definition we can show that the existence of a limit
of a function. Now, same concept we extend for 2 variable functions, now how can you
do that let us see, now we say that for 2 variable functions we say that f approaches
the limit L as x y approaches to x naught y naught. If the values of f(x, y) lie arbitrarily close
to a fixed real numbers L for every x, y sufficiently close to x naught, y naught. You take any
x, y that is sufficiently close to x naught, y naught, f(x) will arbitrarily close to l;
that means, f(x) will approach to L as x,y approaches to x naught, y naught, x, y is
arbitrarily close to x naught, y naught. How can you define this in terms of delta
epsilon let us see, Now here we are having two variable functions instead of a single
variable function. Now, in a single variable functions say we
have x equal to a so, when we take disk at this point at x equal to a. So, this will
be interval, it is a minus delta to a plus delta this is an interval. Now, when we take
point in two variable a function of two variable say here x naught, y naught. Now instead of
a interval it will be a disk centred at this point, it will be a disk centred at x naught,
y naught. So, how can I define limit now, limit x, y tending to x naught, y naught f(x,
y) is equals to L. So, how can we prove whether limit exists and is equal to L.
So, for that again we will repeat the same definition for a two variable functions for
all epsilon greater than 0, there exist corresponding the real number delta greater than 0 such
that mod f (x) minus f(x, y) minus L less than epsilon whenever 0 less than under root
x minus x naught whole square plus y minus y naught whole square less than delta. So,
you take any epsilon greater than 0 there will always exist a corresponding real number
data greater than 0 such that this inequality hold whenever this is less than delta or greater
than zero. So, this is a deleted neighbourhood of x naught, y naught of radius delta so,
this is a basically a circle radius delta. Now, again what does it mean it means that
now this inequality means that f(x, y) is less than L plus epsilon to L minus epsilon
and this means all x, y data lies inside the region inside a disk of radius delta and centre
x naught, y naught; that means, this region centre x naught, y naught and radius delta
this region and this means this is L minus epsilon and this is L plus epsilon and this
is some L. Now, you choose any epsilon greater than 0
there will always exist a corresponding delta such that all those x, y which lie in this
disk, now image of all those x, y is lying in this disk will lie totally in this band
which will lie totally inside this band. No matter how small epsilon or how large epsilon
you are taking they will always exist corresponding delta such that the image of all those x,
y lying in this disk will totally contain in this band.
So, as epsilon tending to 0 this will tends to L and this will tends to x naught y naught
so, this is how we can defined we can defined limit for two variable functions. Now, since
mod of x minus x naught is always less than equals to under root x minus x naught whole
square plus y minus y naught whole square because this is always true.
So, this definition can also be written as mod f(x, y) minus L less than epsilon whenever
0 less than mod x minus x naught less than delta and 0 less than mod y minus y naught
less than delta; that means, instead of disk it may be a rectangle, then also we can apply
the same definition. Now, let us discuss few examples based on this so, that concept of
limit will be more clear. First we have some properties of limit, now
limit, suppose limit x, y tends to x naught, y naught, f(x, y) is L and limit x, y tends
to x naught, y naught, g(x, y) is M, then the addition and subtraction of f(x, y) and
g (x, y) limit x, y tend to x naught, y naught is L plus minus M. And similarly we have the
product of two functions then the limit will also be product. Then scalar multiplication
will be K into L, division f upon g will be simply L upon M. M should not equal to 0 and
similarly we have the next property. These are very straightforward.
Now, come to problems based on delta epsilon definition using delta epsilon definition
show that this is equal to this. The first problem it is limit x, y tending
to 1 comma 2, 2 x plus y is equal to 4. So, how can we prove it, where other way it
is very simple you see when you substitute x equal to 1 and y equal to 2, the value is
2 plus 2 which is 4. Now, if somebody asked how can we prove mathematically that this
limit exists and is equal to 4. So, we have the only option is delta epsilon definition
we can use delta epsilon definition to show that this limit exists and equal to 4, how
can you proceed for that, let epsilon greater than 0 be given.
Now, again we have to show the correspondence of a delta corresponding delta greater than
0 such that we have to show that there exists a corresponding delta greater than 0, such
that mod 2 x plus y less than 4 less than epsilon whenever 0 less than under root x
minus 1 whole square plus y minus 2 whole square less than delta or mod 2 x plus y minus
4 less than epsilon, whenever 0 less than mod x minus 1 less than delta and 0 less than
mod y minus 2 less than delta. So, we can use any definition either this
or this to prove this result now you take mod 2 x plus y minus 4 this is equals to mod
of 2 into x minus 1 plus y minus 2. You can easily see that it is 2 x plus y which is
2 x plus y minus 2 minus 2 is minus 4. Now, this is less than or equals to mod of 2 x
minus 2 x minus 1 plus mod of y minus 2 because mod of a plus b is less than equals to mod
a plus mod b. Now, this is equal to 2 times mod x minus
1 plus y minus 2 now mod x minus 1 so, it will be better if you apply this definition
because this is direct for this particular problem. Now, mod x minus 1 is less than delta
if you take so, this is less than 2 delta and this is again delta which is 3 delta.
So, if we choose or if you take 3 delta less than equal to epsilon then mod of 2 x plus
5 minus 4 less than epsilon whenever 0 less than mod x minus 1 less than delta and 0 less
than mod y minus 2 less than delta. If we choose 3 delta less than equal to epsilon,
then this quantity will be less than epsilon and whenever this is less than delta and this
is less than delta. So, this inequality holds if this delta is
less than equal to epsilon by 3 so, for any epsilon we have shown the corresponding delta
existence of correspond delta such that this inequality holds hence we can say that this
limit exists and is equal to 4. So, the next example, now limit. Now let us
discuss this is simple, now here it is basically a 0 by 0 form so, it is very difficult to
say whether this limit exists or not. Again if this limit exists so, we have to show this
by delta epsilon definition; that means, we have to show the existence of delta corresponding
to every epsilon greater than 0. So, how can you proceed we have supposes limit exist and
we have to show the existence of this limit So, let epsilon greater than 0 be given now
you take x y upon under root x square plus y square so, this is equals to mod of x y
upon under root x square plus y square. Now, we know that x minus y whole square is always
greater than equal to 0 so, x square plus y square minus 2 x y is also greater than
equal 0. So, x y will always be less than or equals to x square plus y square by 2 so,
mod of this quantity will also be less than equals to x square plus y square by 0.
So, this is less than or equals to 1 by 2 x square plus y square upon under root x square
plus y square so, this is equals to under root x square plus y square upon 2. Now, what
we have to show here we have to show that x y upon under root x square minus y square
plus y square minus 0 will be less than epsilon whenever 0 less than under root x square plus
y square less than delta. So, this quantity will be less than delta
by 2 so, if we choose delta by 2 less than equal to epsilon then mod of x y upon under
root x square plus y square will be less than epsilon whenever 0 less than under root x
square plus y square less than delta. So, if you take any delta satisfying this inequality
so, this inequality will be satisfied. So, we have shown that this term to delta corresponding
to any epsilon such that this inequality holds. So, hence we can say that this limit exists
and is equal to 0 so, in this way we can prove that using delta epsilon definition that that
the limit of two or more than two variable function exists and is equal to L. Now, we
will see some more properties of limit of several variable functions in the next lecture.
Thank you very much.
