Hello friends. Welcome to lecture series on
multivariable calculus. In the last lecture, we have seen that; what do you mean by limits
for several variable functions, we have seen that if we write limit x, y tending to x naught,
y naught f (x, y) is equals to L. This means if this limit exist and is equal
to L and it means for every epsilon greater than 0, there exist a corresponding delta
greater than 0 such that such that mod of f (x, y) minus L is less than epsilon whenever
0 less than under root x minus x naught whole square plus y minus y naught whole square
is less than delta. So, we have seen that whatever epsilon may
be no matter how small how large it may be there will always exist a corresponding delta
greater than 0 such that this inequality hold; that means, for every epsilon for every epsilon
there will exist a delta such that a disk centered at x naught y naught of radius delta,
all those x, y lying in that disk will always the image of all those x lying in that disk
will be contained in L minus epsilon 2, L plus epsilon and at the geometric interpretation
of this definition. Now, let us discuss some important properties
of limits the first property is limit x, y tend to x naught, y naught f (x, y) if exist
is always unique ok. Next is to find out the value of the limit the another method is convert
Cartesian coordinate into polar coordinate system; that means, if we substitute x minus
x naught equal to r cos theta y minus y naught equal to r sin theta where r square is equal
to x minus x naught whole square plus y minus y naught whole square and tan theta is equal
to y minus y naught upon x minus x naught that can easily be obtained. If you divide
the second expression y minus y naught is equal to r cos theta and x minus x cos tan
theta, then we obtain tan theta equal is equal to y minus y naught upon x minus x naught
the definition of the limit can be expressed in this way. So, basically if we are having limit x, y
tending to x naught, y naught which we have just explained you f (x, y) equal to L.
Then to find out this limit the another method is to convert Cartesian coordinate into polar
coordinate system. So, how can we do that you simply take x minus x naught as r cos
theta y minus y naught as r sin theta where if you square an add, it is simply r square
is equal to x minus x naught whole square plus y minus y naught whole square and tan
theta is equal to y minus y naught upon x minus x naught. Now as r tend to 0, whatever
theta may be x will tend to x naught and y will tend to y naught that is x, y tent to
x naught, y naught. So; that means, these are there are 2 ways
either you take x , y tend to 0, x naught, y naught tend to 0 and then you can convert
this into polar coordinate system another way out is you simply take x minus x naught
as r cos theta and y minus y naught is r sin theta,. So, now, now this limit will convert
to this will convert to limit r tending to 0 because as r tend to 0, x will tend to x
naught y will tend to y naught that is x, y will tend to x naught, y naught this will
be f (r cos theta, r sin theta) and the limit will be same L.
So, how we can define this in delta epsilon now; so, to show the existence of this limit
again, we will use a concept of delta epsilon that is for every epsilon greater than 0 there
will exist a corresponding delta greater than 0 such that it is mod r less than delta implies
mod of r cos theta r sin theta minus L less than epsilon for all theta for all r and theta,
this must hold for all r and theta. So,so any Cartesian coordinate; if you have
a limit to find out in any Cartesian; so, either you can either you can proceed in the
Cartesian way only or you can convert this into polar coordinate system to find out the
limit. Now, for example, we have this problem changing
into polar coordinate system show that limit of this is equal to 0. Now, let us try this
problem. So, we were discussing about that how we can show existence of a limit by delta
epsilon definition. So, let us call as at this example that is limit x, y tending to
0, 0 x cube upon x square plus y square and a limit is 0. Now, let us try to prove this
that this limit is 0. So, we have 2 ways to show this the first
way is convert x and y into polar coordinate system and the second way is you can proceed
by the usual Cartesian method. So, let us first try to prove it by like converting this
into polar coordinate system. So, we will suppose that x equal to r cos theta and y
is equal to r sin theta. Now as x, y both are tending to 0, 0. So, it will be possible
only when r will tan to 0 ok. So, this limit we will convert into limit r tending to 0
x is r cos theta. So, it is r cube cos cube theta. Now r square
cos square plus r square sin square is r square because cos square plus sin square theta is
1. So, this is equals to limit r tending to 0 it is r cos cube theta and this is clearly
0 for any theta, you can if you take any theta the if limit r is tending to 0. This will
always tend to 0. Now to show this that this is equal to 0, we will again use delta epsilon
definition. So, let epsilon tend to 0 be given. So, it
is mod r cos cube theta minus 0 which is equals to mod r cos cube theta which is equals to
mod r into mod cos cube theta which is less than equals to mod r into 1 because mod cos
theta is always less than equal to 1. So, and this is less than delta. So, if we choose
if we choose delta less than equals to epsilon, then mod of r cos cube theta minus 0 will
be less than epsilon whenever 0 less than mod r less than delta. So, hence we have shown
the existence of such delta for which this inequality hold hence we can say that this
limit exist that is equal to 0. Now, the same can also be proved by the by
the usual like delta epsilon definition without converting this into polar coordinate if we
want to prove this result without using polar coordinate then also we can do that without
using polar coordinate also we can prove this limit. Let epsilon 0 be given. We have to show that mod of x cube upon x
square plus y square minus 0 is less than epsilon whenever 0 less than under root x
minus 0 whole square plus y minus 0 whole square is less than delta this, we have to
prove, we have to prove the existence of such delta.
Now, we take this inequality mod x cube upon x square plus y square minus 0. This is equals
to mod of x into x square upon x square plus y square. This is further equals to mod of
x into mod of x square upon x square plus y square. Now x square is always less than
equals to x square plus y square. So, x square upon x square plus y square is always less
than equal to one and it is also non negative quantity.
So, we can say that it is less than equals to mod x into one and this mod x now you can
use the other definition of, you can use the other definition oflimit or mod x cube upon
x square plus y square or if you want to use same you can use same also.
So, we can use this definition here. So, if you take this is less than delta. So, choose
delta equal to epsilon then we have done. So, we can prove this existence of this limit
without using polar also, but if we use polar coordinate system ah, then we can get the
result easily other properties of limit is to path test for the non existence of a limit
if from 2 different paths as x, y; x, y approaches to x naught, y naught, the function f (x,
y)has different limits, then this implies limit does not exist. So, what does it mean; Let us see. Now they are having x axis, y axis. We have
a point x naught, y naught ok. This point is basically x naught, y naught. We take a
neighbourhood of this point this point x naught, y naught. Now, if you take any neighbourhood
of x naught y naught. All those x, y lying in this region, there are infinite paths by
which this x, y can approach to x naught, y naught, it may be a straight line, it may
be a parabolic curve it may be some other curve ok, there will be infinite paths.
Now, existence of limit means, if we follow any path from x, y to x naught, y naught,
it must be path independent path. Independent means whatever path we follow from x, y to
x naught, y naught, the value of the limit will always be unique if the value of the
limit, if you if you are saying that limit x, y tending to x naught, y naught f (x, y)
is L. This means this means; if you take a neighbourhood
of x naught, y naught and we are taking any x in this neighbourhood, we are infinite paths
from by which x, y can approach to x naught, y naught it must be path independent; that
means, whatever path we follow from x, y to x naught, y naught, the value of this limit
is always L, will always remain the same because it is because limit is unique limit is always
unique if it exist; that means, if from 2 different paths value of the limit are not
same value a double limit we are calling a double limit, if the value of double limit
are not same this means limit does not exist because if limit exist this means it must
be path independent. So, to illustrate this, let us discuss few
examples the first example is limit x, y tending to x naught y naught x, y upon x square plus
y square now 0, 0. Now from x, y to 0, 0, there are infinite
paths we can follow any path, suppose and this is origin and this is any x, y this is
any x, y. So, we can we can move along x axis, we can move along y axis, we can move along
y equal to x, we can move along y equal to 2 x, we can move along y equal to x square
the infinite paths. So, let us move along, let us move along x
axis or y equal to 0. Now if we move along y equal to 0, if you move along y equal to
0 from this point to this point from this path, we are following if you follow this
path, then what is the limit of this expression, there will limit x into 0 y equal to 0 and
x y upon x square plus y square and it is when you substitute x equal to 0 where I substitute
y equal to 0. So, the value is 0. Now, now let us move along
say y axis or x equal to 0. Now if you move along x equal to 0 it is limit y tending to
0 x y upon x square plus y square and x equal to 0 when you substitute x equal to 0 here.
This is 0. Now from these 2 paths value are same; what does it mean from if from 2 paths
2 different paths value of the double limit are same it means the double limit that is
this limit may or may not exist because these are only 2 paths and they are infinite paths
from x, y to 0, 0 the infinite paths. And if from to paths value are same it does
not mean that the value exist value double limits exist and is equal to 0 because there
may be some other path from which the value of this limit may be different for example,
if you take say y equal to x if you take if you move along y equal to x along y equal
to x. Then limit x tending to 0; you substitute
y equal to x it is x square upon x square plus x square which is limit x tending to
0; x square upon 2 x square which is 1 by 2. Now from this path from this path value
is 0 from this path value is 0 and from some other path value is 1 by 2.
So, values are not same values are different this means this limit does not exist. So,
we can simply say this implies limit x, y tending to 0, 0 x y upon x square plus y square.
It does not exist, why does not exist because from 2 different paths values are different
now the other way out to show that limit does not exist is other way out is you take you
take a path general path you take a general path you move along say y equals to m x. If you move along y equal to m x, this means
it is limit, you substitute y equal to m x it is x into m x upon x square plus m x whole
square and x is tending to 0. Remember this that this path must pass through
x naught, y naught must pass here x naught, y naught is 0, 0. So, this path must pass
through as 0, 0 whatever path we are choosing it must must pass through this point now this
is equals to limit x tending to 0, it is m x square upon x square times one plus m square
x square cancel out and it is m upon one plus m square.
Now this value the limit this value comes out to be dependent on m you take different
values of m say you take m equal to 1. This value is 1 by 2 you take m equal to 2, then
this value is 2 upon 5. So, for different values of m the value of
the limit are different this means limit does not exist because now it is path dependent
we take different paths values are different it is path dependent; however, it must if
limit exist it must be path independent. So, it depends on m
this implies limit x y tend to 0, 0; x y upon x square plus y square, it does not exist.
So, basically to show to show that limit does not exist the double limit does not exist.
We have 2 ways the first way is you take 2 different paths and try to show that from
2 different paths value of the limit are different. The other way out is you try to show that
it is path dependent you take some arbitrary path like y equal to m x or y equal to k x
square or something and try to show that it is it depends on m or k in this way we can
show that limit does not exist. Say we have second example it is limit x,
y tending to 0, 0 the problem is x cube y upon x raise to the power 6 plus y raise to
the power 6. Now if we move along, if we move along say y equal to 0, if you move along
y equal to 0, then this value when you substitute y equal to 0 then this is clearly 0 because
it when you substitute y equal to 0 here this is 0. Now you move along say y equal to x
cube around this curve. If you move along y equal to x cube, then
this is nothing, but limit x tending to 0; x cube into x cube upon x to power 6 plus
it is x to the power 6 which is equal to limit x tending to 0 x to the power 6 upon 2 into
x raise to the power 6; which is 1 by 2. So, from one path value is 0 and from other path
value is 1 by 2 this means this limit does not exist. Now the next example next example is limit
x, y, z tending to 0, 0, 0, it is x y z upon x square plus y to the power 4 plus z to the
power 4. Now, we have to find a path such that it comes out of a path dependent to show
that this limit does not exist. So, we can choose some path say, we can take, let x is
equals to some k t square say y equal to y equal to say t and z equal to t where t is
some parameter basically in 3 d, we are taking this curve.
Now, is substitute this it is limit x is k t square y is t and z is t and it is k square
p raised to power 4 plus t raised to power 4 plus t raised to power 4 and limit t tends
to 0 because as x y z all tend to 0 this will happen only when t with t tending to 0 and
this is equals to limit t tending to 0. It is k into t raised to power 4 upon k plus
k a square plus 2. So, this will be equal to k of k upon k square plus 2 that is depends
on k it depends on k this means, this means this limit does not exist.
So, in this way we can show that double limit does not exist now if you take say if you
take 4 or 5 paths and the value of the limit always come out to the same, then again this
does not guarantee that the limit exist because there may be some other path by which the
value of the limit comes out to be different, if we have to show the existence of a limit
we have only option is delta epsilon definition, we have to show the existence of a limit using
delta epsilon definition only to show that the limit does not exist, we can use this
concept we can we can use 2 different path and try to show the value of limit comes out
to be different or where we can try to show that it is path dependent. Now, we will talk about iterated limits and
double limit now what does it mean you see that double limit is this thing. This is x,
y tending to x naught, y naught f( x y), suppose it exist and equal to L and we have it is
it is called double limit also called the double limit and other things are iterated
limit iterated limit means limit x tend to x naught limit y tend to y naught f x y or
limit y tend to y naught limit x tend to x naught, f (x y), these are called iterated
limits. Now, now if you take now if you take x naught,
y naught here and you take a neighbourhood of this point centroid x naught y naught,
you take any x y in this disk this means you first you first stage y tend to y naught keeping
x constant and then you take x x tend to x naught. So, first you are taking y tend to
y naught this is this is this is x naught y naught first you are taking y tending to
y naught means this thing y tend into y naught, this is some point x, y ; y tend to y naught
now this now this point is this point is x naught y naught now here first x tend to x
naught and then y tend to y naught. So, we come to this point.
So, this is y tend to y naught and then x tend to x naught. So, we come to this point.
So, these are these are basically 2 different paths, one path is this another path is this.
Now if this limit exist and is equal to L, then the iterated limit, then the iterated
limit value the iterated limit is also equal to L provided limit y tending to y naught
f (x, y) and limit tending to x naught f (x, y), exist if this condition hold, then only
we can say that if double limit exist, then iterated limit also exist equal to L. So, basically if this is equal to L, then
this implies, then this condition implies if this is equal to L that is then this condition
implies that these are also equal and is equal to L provided.
Provided this inside limit exist because if this limit exist then these are
basically 2 paths if this limit exist and these are basically 2 paths and if this is
equal to L this means it is path independent if it is path independent then from these
2 paths also the value will be same, value will be L now if we see the converse part
if this is if this exist and suppose this and this are equal to L, then these are 2
only 2 paths ok, if this and if this and this limit exist then these are only 2 paths and
from these 2 paths if limit comes out to be L then this double limit may or may not exist
because basically these iterated limit if this limit exist are only 2 paths.
So, let us understand this by giving some examples you see suppose you take this
it is x plus y upon x minus y. Suppose you want to compute, this limit x
minus y should not equal to see. Now if you find this limit x tend to 0, limit y tending
to 0 x plus y upon x minus y if you find this limit, this is this iterated limit, then this
is limit x tend to 0 you simply substitute you simply tend to 0 then it is x plus 0 upon
x minus 0 and when you take x into 0 then this is one.
Now, you take the other iterated limit; it is limit y tend to 0, it is 0 plus y upon
0 minus y and when you take y tend to 0 it is minus one. So, iterated limit exist and
are not equal you see you see that this, this limit and this limit exist this limit is 1
and this limit is minus 1, this and this limit exist and iterated limit are not same this
means this implies limit x, y tend to 0, 0 x plus y upon x minus y, this does not exist
because if because if this inside limit exist, then this iterated limiter simply 2 paths
and from the 2 different paths values are different this means this limit double limit
does not exist now see another example it is limit x, y tending to 0, 0. It is x square y square upon it is again x
square y square plus x minus y whole to square ok, the problem is find a double limit and
the iterated limit if they exist provided denominator is not equal to 0.
Now, first you find the double limits ok. So, you take limit y tending to 0 limit x
tend to 0, f(x,y) which is x square y square upon x square y square plus x minus y whole
square. Now when you put x when you take x tend to 0, here in this expression. So, this
will tend to 0, then this is simply equal to 0 ok, one can easily see that when you
take x tend to 0 in this expression. So, numerator is 0. So, the value is 0 now the other iterated
limit is limit x tend to 0 suppose and limit y tend to 0 x square y square upon x square
y square plus x minus y whole square. Now, when you take y tend to 0 again numerator
is zero. So, this value is again 0 now these this inside limits exist and the iterated
limits are same what does it mean ? What can we say about double limit from here, we can
say that double limit may or may not exist because these are only 2 paths it may possible
from some other path value are double limit comes out to be different from this limit
from this value say if you take a path say you take a path along say take a path y equal
to x, if you take a path y equal to x a here. So, we will obtain limit x tend to 0, x square
x square upon x is to power 4 plus 0 which is one from this path they are getting value
one and from other paths they are getting a value 0.
This means limit does not exist because there are 2 different paths from which value of
the limits are different though this means a path dependent then this implies this limit does not exist. So, hence we can easily show
get whether a limit exist or it does not exist to show the existence we have to go only through
delta epsilon definition to show that the limit does not exist we have to we have to
show that from 2 different paths values of the limit are different.
Thank you very much.
