WEBVTT
Kind: captions
Language: en
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Hello friends. Welcome to lecture series on
multivariable calculus. In the last lecture,
00:00:22.770 --> 00:00:29.880
we have seen that; what do you mean by limits
for several variable functions we have seen
00:00:29.880 --> 00:00:44.210
that if we write limit x y tending to x naught
y naught f x y is equals to L.
00:00:44.210 --> 00:00:54.020
This means if this limit exist and is equal
to L and it means for every epsilon greater
00:00:54.020 --> 00:01:11.600
than 0, there exist a corresponding delta
greater than 0 such that such that mod of
00:01:11.600 --> 00:01:26.200
f x y minus L is less than epsilon whenever
whenever 0 less than under root x minus
00:01:26.200 --> 00:01:34.220
x naught whole square plus y minus y naught
whole square is less than delta .
00:01:34.220 --> 00:01:41.189
So, we have seen that whatever epsilon may
be no matter how small how large it may be
00:01:41.189 --> 00:01:47.740
there will always exist a corresponding delta
greater than 0 such that this inequality hold;
00:01:47.740 --> 00:01:55.130
that means, for every epsilon for every epsilon
there will exist a delta such that a disk
00:01:55.130 --> 00:02:02.150
centered at x naught y naught of radius delta,
all those x y lying in that disk will always
00:02:02.150 --> 00:02:09.340
the image of all those x lying in that disk
will be contained in L minus epsilon 2 L plus
00:02:09.340 --> 00:02:13.450
epsilon and at the geometric interpretation
of this definition .
00:02:13.450 --> 00:02:19.769
Now, let us discuss some important properties
of limits the first property is limit x y
00:02:19.769 --> 00:02:27.799
tan to x naught y naught f x y if exist is
always unique ok. Next is to find out the
00:02:27.799 --> 00:02:33.160
value of the limit the another method is convert
Cartesian coordinate into polar coordinate
00:02:33.160 --> 00:02:40.280
system; that means, if we substitute x minus
x naught equal to r cos theta y minus y naught
00:02:40.280 --> 00:02:46.250
equal to r sin theta where r square is equal
to x minus x naught whole square plus y minus
00:02:46.250 --> 00:02:51.709
y naught whole square and tan theta is equal
to y minus y naught upon x minus x naught
00:02:51.709 --> 00:02:57.660
that can easily be obtained. If you divide
the second expression y minus y naught
00:02:57.660 --> 00:03:04.260
is equal to r cos theta and x minus x cos
tan theta, then we obtain tan theta equal
00:03:04.260 --> 00:03:09.709
is equal to y minus y naught upon x minus
x naught the definition of the limit can be
00:03:09.709 --> 00:03:10.709
expressed in this way .
00:03:10.709 --> 00:03:18.990
So, basically if we are having limit x y tending
to x naught y naught which we have just explained
00:03:18.990 --> 00:03:29.330
you f x y equal to L .
Then to find out this limit the another method
00:03:29.330 --> 00:03:34.820
is converges Cartesian coordinate into polar
coordinate system . So, how can we do that
00:03:34.820 --> 00:03:47.299
you simply take x minus x naught as r cos
theta y minus y naught as r sin theta where
00:03:47.299 --> 00:03:54.090
if you square an add, it is simply r square
is equal to x minus x naught whole square
00:03:54.090 --> 00:04:03.180
plus y minus y naught whole square and tan
theta is equal to y minus y naught upon x
00:04:03.180 --> 00:04:14.010
minus x naught ok . Now as r tan into 0 whatever
theta may be x will tend to x naught and y
00:04:14.010 --> 00:04:19.459
will tend to y naught that is x y tan tan
x y will tend to x naught y naught ok .
00:04:19.459 --> 00:04:26.360
So; that means, these are there are 2 ways
either you convert x by tan to 0 x naught
00:04:26.360 --> 00:04:31.090
y naught to x y tan to 0 0 and then you can
convert this into polar coordinate system
00:04:31.090 --> 00:04:36.979
another way out is you simply take x minus
x naught as r cos theta and y minus y naught
00:04:36.979 --> 00:04:44.439
is r sin theta, ok . So, now, now this limit
will convert to this will convert to limit
00:04:44.439 --> 00:04:49.900
r tending to 0 because as r tend to 0 will
tend to x naught y will tend to y naught that
00:04:49.900 --> 00:04:56.979
is x y will tend to x naught y naught ok this
will be f of r cos theta r sin theta and the
00:04:56.979 --> 00:05:01.110
limit will be same L .
So, how we can define this in delta epsilon
00:05:01.110 --> 00:05:05.849
now; so, to show the existence of this limit
again, we will use a concept of delta epsilon
00:05:05.849 --> 00:05:15.479
that is for every epsilon greater than 0 there
will exist a corresponding delta greater than
00:05:15.479 --> 00:05:33.460
0 such that it is mod r less than delta implies
mod of r cos theta r sin theta minus L less
00:05:33.460 --> 00:05:43.479
than epsilon for all theta for all r and theta,
this must hold for all are for all r and theta
00:05:43.479 --> 00:05:48.759
ok .
So so, so any Cartesian coordinate;
00:05:48.759 --> 00:05:55.159
if you have a limit to find out in any
Cartesian ; so, either you can either you
00:05:55.159 --> 00:05:59.509
can proceed in the Cartesian way only or you
can convert this into polar coordinate system
00:05:59.509 --> 00:06:04.879
to find out the limit .
Now, for example, we have this problem
00:06:04.879 --> 00:06:11.360
changing into polar coordinate system show
that limit of this is equal to 0 . Now, let
00:06:11.360 --> 00:06:17.610
us try this problem . So, we were discussing
about that how we can show existence of
00:06:17.610 --> 00:06:24.379
a limit by delta epsilon definition . So,
let us call as at this example that is limit
00:06:24.379 --> 00:06:37.330
x y tending to 0 0 x cube upon x square plus
y square and a limit is 0 ok . Now, let us
00:06:37.330 --> 00:06:43.009
try to prove this that this limit is 0 .
So, we have 2 ways to show this the first
00:06:43.009 --> 00:06:51.949
way is convert x and y into polar coordinate
system ok and the second way is you can
00:06:51.949 --> 00:06:59.770
proceed by the usual Cartesian method ok
. So, let us first try to prove it by like
00:06:59.770 --> 00:07:07.159
converting this into polar coordinate system
. So, we will suppose that x equal to r cos
00:07:07.159 --> 00:07:17.090
theta and y is equal to r sin theta . Now
as x y both are tending to 0 0 . So, it will
00:07:17.090 --> 00:07:25.069
be possible only when r will tan to 0 ok.
So, this limit we will convert into limit
00:07:25.069 --> 00:07:34.169
r tending to 0 x is r cos theta.
So, it is r cube cos cube theta . Now r square
00:07:34.169 --> 00:07:41.069
cos square plus r square sin square is r square
because cos square plus sin square theta is
00:07:41.069 --> 00:07:50.840
1 . So, this is equals to limit r tending
to 0 it is r cos cube theta and this is clearly
00:07:50.840 --> 00:08:01.939
0 for any theta , you can if you take any
theta the if limit r is tending to 0. This
00:08:01.939 --> 00:08:07.860
will always tend to 0 . Now to show this that
this is equal to 0, we will again use delta
00:08:07.860 --> 00:08:18.779
epsilon definition ok .
So, let epsilon get tan 0 be given ok . So,
00:08:18.779 --> 00:08:28.009
it is mod r cos cube theta minus 0 which is
equals to mod r cos cube theta which is equals
00:08:28.009 --> 00:08:37.110
to mod r into mod cos cube theta which is
less than equals to mod r into 1 because mod
00:08:37.110 --> 00:08:48.830
cos theta is always less than equal to 1 . So,
and this is less than delta . So, if we choose
00:08:48.830 --> 00:08:58.610
if we choose delta less than equals to epsilon
, then mod of r cos cube theta minus 0 will
00:08:58.610 --> 00:09:09.399
be less than epsilon whenever 0 less than
mod r less than delta . So, hence we have
00:09:09.399 --> 00:09:17.100
shown the existence of such delta for which
this inequality hold hence we can say that
00:09:17.100 --> 00:09:24.430
this limit exist that is equal to 0 ok .
Now, the same can also be proved by by the
00:09:24.430 --> 00:09:29.680
by the usual like delta epsilon definition
without converting this into polar coordinate
00:09:29.680 --> 00:09:35.370
if we want to prove this result without using
polar coordinate then also we can do that
00:09:35.370 --> 00:09:40.390
without using polar coordinate also we can
prove this limit. Let delta let epsilon 0
00:09:40.390 --> 00:09:47.090
be given .
We have to show that mod of x cube upon
00:09:47.090 --> 00:09:59.910
x square plus y square minus 0 is less than
epsilon whenever 0 less than under root
00:09:59.910 --> 00:10:08.440
x minus 0 whole square plus y minus 0 whole
square is less than delta this, we have to
00:10:08.440 --> 00:10:13.220
prove , we have to prove that distance of
such delta ok .
00:10:13.220 --> 00:10:21.600
Now, we take this inequality mod x cube upon
x square plus y square minus 0 . This is equals
00:10:21.600 --> 00:10:30.980
to mod of x into x square upon x square plus
y square . This is further equals to mod of
00:10:30.980 --> 00:10:40.050
x into mod of x square upon x square plus
y square . Now x square is always less than
00:10:40.050 --> 00:10:47.089
equals to x square plus y square ok . So,
x square upon x square plus y square is always
00:10:47.089 --> 00:10:51.060
less than equal to one and it is also non
negative quantity .
00:10:51.060 --> 00:11:01.000
So, we can say that it is less than equals
to mod x into one ok and this mod x now you
00:11:01.000 --> 00:11:10.959
can use the other definition of , you can
use the other definition of limit or
00:11:10.959 --> 00:11:18.570
mod x cube upon x square plus y square or
if you want to use same you can use same
00:11:18.570 --> 00:11:34.910
also .
So, we can use this definition here . So,
00:11:34.910 --> 00:11:42.320
if you take this is less than delta . So,
choose delta equal to epsilon then we have
00:11:42.320 --> 00:11:50.389
done . So, we can prove this existence
of this limit without using polar also , but
00:11:50.389 --> 00:11:57.590
if we use polar coordinate system then
we can get the result easily ok other properties
00:11:57.590 --> 00:12:05.970
of limit limit is to path test for the
non existence of a limit if from 2 different
00:12:05.970 --> 00:12:11.440
paths as x y; x y approaches to x naught y
naught , the function f x y has different
00:12:11.440 --> 00:12:16.779
limits, then this implies limit does not exist
ok .
00:12:16.779 --> 00:12:23.820
So, what does it mean; Let us see .
Now they are having x axis, y axis . We have
00:12:23.820 --> 00:12:31.269
a point x naught y naught ok. This point is
basically x naught y naught . We take a neighbourhood
00:12:31.269 --> 00:12:38.709
of this point this point x naught y naught
ok . Now, if you take any neighbourhood of
00:12:38.709 --> 00:12:46.639
x naught y naught ok . All those x y lying
in this region , there are infinite paths
00:12:46.639 --> 00:12:51.589
by which this x y can approach to x naught
y naught, it may be a straight line, it may
00:12:51.589 --> 00:12:59.010
be a parabolic curve it may be some other
curve ok, there will be infinite paths ok
00:12:59.010 --> 00:13:04.610
.
Now, existence of limit means , if we follow
00:13:04.610 --> 00:13:12.700
any path from x y to x naught y naught , it
must be path independent path. Independent
00:13:12.700 --> 00:13:20.840
means whatever path we follow from x y 2 x
naught y naught , the value of the limit will
00:13:20.840 --> 00:13:26.500
always be unique if the if the value of the
limit , if you if you are saying that limit
00:13:26.500 --> 00:13:36.980
x y tending to x naught y naught f x y is
L ok .
00:13:36.980 --> 00:13:43.100
This means this means; if you take a neighbourhood
of x naught y naught and we are taking any
00:13:43.100 --> 00:13:48.660
x in this neighbourhood , we are infinite
paths from by which x y can approach to
00:13:48.660 --> 00:13:55.560
x naught y naught it must be path independent;
that means, whatever path we follow from x
00:13:55.560 --> 00:14:00.190
y to x naught y naught , the value of this
limit is always; L will always remain the
00:14:00.190 --> 00:14:08.920
same because it is because limit is unique
limit is always unique if it exist ; that
00:14:08.920 --> 00:14:16.870
means, if from 2 different paths value of
the limit are not same value a double limit
00:14:16.870 --> 00:14:23.899
we are calling a double limit , if the value
of double limit are not same this means limit
00:14:23.899 --> 00:14:29.820
does not exist because if limit exist this
means it must be path independent .
00:14:29.820 --> 00:14:40.779
So, to illustrate this, let us discuss few
examples ok the first example is limit x y
00:14:40.779 --> 00:14:55.459
tending to x naught y naught x y upon x square
plus y square now ok 0 0 it is 0 0 .
00:14:55.459 --> 00:15:06.560
Now from x y to 0 0, they are infinite paths
ok we can follow any path , suppose and this
00:15:06.560 --> 00:15:13.980
is origin ok and this is any x y this is any
x y . So, we can we can move along x axis,
00:15:13.980 --> 00:15:18.839
we can move along y axis, we can move along
y equal to x, we can move along y equal to
00:15:18.839 --> 00:15:23.329
2 x, we can move along y equal to x square
the infinite paths ok .
00:15:23.329 --> 00:15:35.819
So, let us move along , let us move along
x axis or y equal to 0 . Now if we move along
00:15:35.819 --> 00:15:42.300
y equal to 0 , if you move along y equal to
0 from this point to this point from this
00:15:42.300 --> 00:15:48.269
path, we are following if you follow this
path , then what is the limit of this expression
00:15:48.269 --> 00:15:57.311
, there will limit x into 0 y equal to 0 ok
and x y upon x square plus y square and it
00:15:57.311 --> 00:16:01.490
is when you substitute x equal to 0 where
I substitute y equal to 0.
00:16:01.490 --> 00:16:15.430
So, the value is 0 ok . Now, now let us move
along along say y axis or x equal to 0 . Now
00:16:15.430 --> 00:16:26.639
if you move along x equal to 0 it is limit
y tending to 0 x y upon x square plus y square
00:16:26.639 --> 00:16:34.629
and x equal to 0 when you substitute x equal
to 0 here. This is 0 . Now from these 2 paths
00:16:34.629 --> 00:16:44.110
value are same; what does it mean from if
if from 2 paths 2 different paths value of
00:16:44.110 --> 00:16:51.100
the double limit are same it means the double
limit that is this limit may or may not exist
00:16:51.100 --> 00:16:57.420
because these are only 2 paths and they are
infinite paths from x y to 0 0 the infinite
00:16:57.420 --> 00:17:03.079
paths ok .
And if from to paths value are same it does
00:17:03.079 --> 00:17:07.980
not mean that the value exist value double
limits exist and is equal to 0 because there
00:17:07.980 --> 00:17:14.910
there may be some other path from which the
value of this limit may be different for example,
00:17:14.910 --> 00:17:24.570
if you take say y equal to x if you take if
you move along y equal to x along y equal
00:17:24.570 --> 00:17:25.930
to x .
00:17:25.930 --> 00:17:36.320
Then limit x into 0 you substitute y equal
to x it is x square upon x square plus x square
00:17:36.320 --> 00:17:47.200
which is limit x into 0 x square upon 2 x
square which is 1 by 2 . Now from this path
00:17:47.200 --> 00:17:52.900
from this path value is 0 from this path value
is 0 and from some other path value is 1 by
00:17:52.900 --> 00:17:55.900
2.
So, values are not same values are different
00:17:55.900 --> 00:18:05.250
this means this limit does not exist . So,
we can simply say this implies limit x y tending
00:18:05.250 --> 00:18:22.300
to 0 0 x y upon x square plus y square . It
does not exist ok , why does not exist because
00:18:22.300 --> 00:18:29.540
from 2 different paths values are different
ok now the other way out to show that limit
00:18:29.540 --> 00:18:37.720
does not exist is other way out is you take
you take a path general path ok you take a
00:18:37.720 --> 00:18:44.290
general path you move along say y equals to
m x .
00:18:44.290 --> 00:18:49.630
If you move along y equal to m x, this means
it is limit , you substitute y equal to m
00:18:49.630 --> 00:19:00.930
x it is x into m x upon x square plus m x
whole square and x is tending to 0 .
00:19:00.930 --> 00:19:05.950
Remember this that this path must path through
x naught y naught must path here x naught
00:19:05.950 --> 00:19:13.840
y naught is 0 0 . So, this path must path
through as 0 0 ok whatever path we are choosing
00:19:13.840 --> 00:19:22.260
it must must path through this point now this
is equals to limit x into 0 , it is m x square
00:19:22.260 --> 00:19:31.799
upon x square times one plus m square x square
cancel out and it is m upon one plus m square
00:19:31.799 --> 00:19:36.200
.
Now this value the limit this value comes
00:19:36.200 --> 00:19:42.581
out to be dependent on m you take different
values of m say you take m equal to 1. This
00:19:42.581 --> 00:19:47.930
value is 1 by 2 you take m equal to 2, then
this value is 2 upon 5 .
00:19:47.930 --> 00:19:53.260
So, for different values of m the value of
the limit are different this means limit does
00:19:53.260 --> 00:20:00.560
not exist because now it is path dependent
we take different paths values are different
00:20:00.560 --> 00:20:05.790
it is path dependent ; however, it must if
limit exist it must be path independent .
00:20:05.790 --> 00:20:23.950
So, it depends on m
this implies limit x y tend to 0 0 x y upon
00:20:23.950 --> 00:20:35.340
x square plus y square , it does not exist
. So, basically 2 showed to show that limit
00:20:35.340 --> 00:20:41.670
does not exist the double limit does not exist.
We have 2 ways the first way is you take 2
00:20:41.670 --> 00:20:47.720
different paths and try to show that from
2 different paths value of the limit are different
00:20:47.720 --> 00:20:55.160
. The other way out is you try to show that
it is path dependent you take some arbitrary
00:20:55.160 --> 00:21:00.970
path like y equal to m x or y equal to k x
square or something and try to show that it
00:21:00.970 --> 00:21:09.240
is it depends on m or k ok in this way we
can show that limit does not exist .
00:21:09.240 --> 00:21:26.900
Say we have second example it is limit x y
tending to 0 0 the problem is x cube y upon
00:21:26.900 --> 00:21:37.520
ok . Now if we move along , if we move along
say y equal to 0 , if you move along y equal
00:21:37.520 --> 00:21:45.340
to 0, then this value when you substitute
y equal to 0 then this is clearly 0 ok because
00:21:45.340 --> 00:21:54.500
it when you substitute y equal to 0 here this
is 0 . Now you move along say y equal to x
00:21:54.500 --> 00:22:01.020
cube around this curve .
If you move along y equal to x cube, then
00:22:01.020 --> 00:22:13.580
this is nothing, but limit x tending to 0
x cube into x cube upon x to power 6 plus
00:22:13.580 --> 00:22:23.900
it is x to the power 6 which is equal to limit
x tending to 0 x to the power 6 upon which
00:22:23.900 --> 00:22:30.420
is 1 by 2 . So, from one path value is 0 and
from other path value is 1 by 2 this means
00:22:30.420 --> 00:22:38.930
this limit does not exist .
Now the next example next example is limit
00:22:38.930 --> 00:22:52.760
x y z tending to 0 0 0 , it is x y z x square
plus it is right to the power 4 plus z to
00:22:52.760 --> 00:23:03.220
the power 4 . Now, we have to find a path
such that it comes out of a path dependent
00:23:03.220 --> 00:23:09.850
to show that this limit does not exist ok
. So, we can choose some path say, we can
00:23:09.850 --> 00:23:24.750
take , let x is equals to some k t square
say y equal to y equal to say t and z equal
00:23:24.750 --> 00:23:34.220
to t where t is some parameter ok basically
in in 3 d, we are taking this curve
00:23:34.220 --> 00:23:40.470
ok .
Now, is substitute this it is limit x is k
00:23:40.470 --> 00:23:49.730
t square y is t and z is t and it is k square
p raised to power 4 plus t raised to power
00:23:49.730 --> 00:23:57.390
4 plus t raised to power 4 and limit t tends
to 0 because because as x y z all tend to
00:23:57.390 --> 00:24:05.980
0 this will happen only when t with t tending
to 0 ok and this is equals to limit t tending
00:24:05.980 --> 00:24:17.370
to 0. It is k into t raised to power 4 upon
upon k plus k a square plus 2 . So, this
00:24:17.370 --> 00:24:26.150
will be equal to k of k upon k square plus
2 that is depends on k it depends on k this
00:24:26.150 --> 00:24:42.710
means , this means this limit
00:24:42.710 --> 00:24:51.070
does not exist
ok .
00:24:51.070 --> 00:24:58.670
So, in this way we can show that double limit
does not exist now if you take say if you
00:24:58.670 --> 00:25:05.590
take 4 or 5 paths and the value of the limit
always come out to the same , then again this
00:25:05.590 --> 00:25:13.650
does not guarantee that the limit exist because
there may be some some other path by which
00:25:13.650 --> 00:25:20.270
the value of the limit comes out to be different
, if we have to show the existence of a limit
00:25:20.270 --> 00:25:27.020
we have only option is delta epsilon definition,
we have to show the existence of a limit using
00:25:27.020 --> 00:25:34.620
delta epsilon definition only to show that
the limit does not exist, we can use this
00:25:34.620 --> 00:25:40.900
this concept we can we can use 2 different
path and try to show the value of limit comes
00:25:40.900 --> 00:25:46.000
out to be different or where we can try to
show that it is path dependent .
00:25:46.000 --> 00:25:51.970
Now, we will talk about iterated limits and
double limit now what does it mean you see
00:25:51.970 --> 00:26:00.480
that double limit is this thing. This is x
y tending to x naught y naught f x y , suppose
00:26:00.480 --> 00:26:07.851
it exist and equal to L ok and we have it
is it is called double limit also called the
00:26:07.851 --> 00:26:17.210
double limit ok and other things are iterated
limit iterated limit means limit x tend to
00:26:17.210 --> 00:26:34.770
x naught limit y tend to y naught f x y or
limit y tend to y naught limit x to x naught
00:26:34.770 --> 00:26:44.870
f x y , these are called iterated limits
ok .
00:26:44.870 --> 00:26:58.400
Now, now if you take now if you take x naught
y naught here ok and you take a neighbourhood
00:26:58.400 --> 00:27:07.990
of this point centroid x naught y naught , you
take any x y in this disk this means you first
00:27:07.990 --> 00:27:13.130
you first stage y tend to y naught keeping
x constant and then you take x x tan to x
00:27:13.130 --> 00:27:19.550
naught . So, first you are taking y tend to
y naught this is this is this is x naught
00:27:19.550 --> 00:27:25.830
y naught ok first you are taking y tending
to y naught means this thing ok y tend into
00:27:25.830 --> 00:27:34.630
y naught ok , this is some point x y ok y
tend to y naught now this now this point is
00:27:34.630 --> 00:27:40.911
this point is x naught y naught ok now
here first x tend to x naught and then y tan
00:27:40.911 --> 00:27:44.770
to y naught. So, we come to this point ok
.
00:27:44.770 --> 00:27:56.330
So, this is y tend to y naught and then x
tend to x naught. So, we come to this point
00:27:56.330 --> 00:28:06.160
ok ok . So, t these are these are basically
2 different paths, one path is this another
00:28:06.160 --> 00:28:17.880
path is this ok . Now now if this limit
exist and is equal to L , then the iterated
00:28:17.880 --> 00:28:25.750
limit, then the iterated limit value the iterated
limit is also equal to L provided provided
00:28:25.750 --> 00:28:31.780
limit y tending to y naught f x y and limit
actually as x naught f x y.
00:28:31.780 --> 00:28:38.240
Exist if this condition hold , then only we
can say that if double limit exist, then
00:28:38.240 --> 00:28:43.760
iterated limit also exist equal to l.
So, basically if this is equal to L, then
00:28:43.760 --> 00:28:50.190
this implies, , then this condition implies
if this is equal to L that is then this condition
00:28:50.190 --> 00:28:58.440
implies that these are also equal and is equal
to L provided .
00:28:58.440 --> 00:29:16.160
Provided this inside limit exist
00:29:16.160 --> 00:29:26.660
because because if this limit exist then these
are basically 2 paths ok if this limit exist
00:29:26.660 --> 00:29:32.031
and these are basically 2 paths and if this
is equal to L this means it is path independent
00:29:32.031 --> 00:29:38.000
if it is path independent then from these
2 paths also the value will be same value
00:29:38.000 --> 00:29:45.580
will be L now if we see the converse path
if this is if this exist and suppose this
00:29:45.580 --> 00:29:54.970
and this are equal to L , then then these
are 2 only 2 paths ok, if this and if this
00:29:54.970 --> 00:30:00.510
and this limit exist then these are only 2
paths and from these 2 paths if limit comes
00:30:00.510 --> 00:30:08.520
out to be L then this double limit may or
may not exist because basically these iterated
00:30:08.520 --> 00:30:14.280
limit if this limit exist are only 2 paths
ok .
00:30:14.280 --> 00:30:28.490
So, let us understand this by giving some
examples you see suppose you take this
00:30:28.490 --> 00:30:39.030
it is x plus y upon x minus y .
Suppose you want to compute, this this limit
00:30:39.030 --> 00:30:51.950
x minus y should not equal to see . Now now
if you find this limit limit x tend to 0 , limit
00:30:51.950 --> 00:30:59.240
y tending to 0 x plus y upon x minus y if
you find this limit, this is this iterated
00:30:59.240 --> 00:31:05.950
limit, then this is limit x tend to 0 you
simply substitute you simply tan by tend to
00:31:05.950 --> 00:31:15.860
0 then it is x plus 0 upon x minus 0 and when
you take x into 0 then this is one ok .
00:31:15.860 --> 00:31:34.450
Now, you take the other iterated limit
it is limit y tend to 0 it is 0 plus y upon
00:31:34.450 --> 00:31:44.120
0 minus y and when you take y tend to 0 it
is minus one . So, iterated limit exist and
00:31:44.120 --> 00:31:54.420
are not equal ok you see you see that this,
this limit and this limit exist this limit
00:31:54.420 --> 00:32:01.310
is 1 and this limit is minus 1, this and this
limit exist ok and iterated limit are not
00:32:01.310 --> 00:32:15.530
same this means this implies limit x y tend
to 0 0 x plus y upon x minus y , this does
00:32:15.530 --> 00:32:26.610
not exist
ok because if because if this inside limit
00:32:26.610 --> 00:32:33.520
exist , then this iterated limiter simply
2 paths and from the 2 different paths values
00:32:33.520 --> 00:32:41.760
are different this means this limit double
limit does not exist ok now see another example
00:32:41.760 --> 00:32:57.400
it is limit x y tending to 0 0 .
It is x square y square upon it is again x
00:32:57.400 --> 00:33:07.460
square y square plus x minus y whole to square
ok, the problem is find a double limit and
00:33:07.460 --> 00:33:13.240
the iterated limit if they exist the provided
denominator is not equal to 0 .
00:33:13.240 --> 00:33:26.230
Now, first you find the double limits ok.
So, you take limit y tending to 0 limit x
00:33:26.230 --> 00:33:33.520
tend to 0 f x y which is x square y square
upon x square y square plus x minus y whole
00:33:33.520 --> 00:33:39.940
square . Now when you put x when you take
x x tend to 0, here in this expression. So,
00:33:39.940 --> 00:33:46.530
this this will tend to 0, then this is simply
equal to 0 ok, one can easily see that when
00:33:46.530 --> 00:33:52.510
you take x tend to 0 in this expression. So,
numerator is 0. So, the value is 0 now the
00:33:52.510 --> 00:34:01.810
other iterated limit is limit x tend to 0
suppose and limit y tend to 0 x square y square
00:34:01.810 --> 00:34:05.230
upon x square y square plus x minus y whole
square .
00:34:05.230 --> 00:34:15.810
Now, when you take y tend to 0 again numerator
is zero. So, this value is again 0 now these
00:34:15.810 --> 00:34:22.480
this inside limits exist and the iterated
limits are same what does it mean ? What can
00:34:22.480 --> 00:34:28.889
we say about double limit from here, we can
say that double limit may or may not exist
00:34:28.889 --> 00:34:35.379
because these are only 2 paths it may possible
from from some other path value are double
00:34:35.379 --> 00:34:41.149
limit comes out to be different from this
limit from this value say say if you take
00:34:41.149 --> 00:34:49.740
a path say you take a path along say take
a path y equal to x , if you take a path y
00:34:49.740 --> 00:34:59.130
equal to x a here. So, we will obtain limit
x tend to 0 x square x square upon x is to
00:34:59.130 --> 00:35:05.430
power 4 plus 0 which is one from this path
they are getting value one and from other
00:35:05.430 --> 00:35:09.779
paths they are getting a value 0.
This means limit does not exist because there
00:35:09.779 --> 00:35:15.990
are 2 different paths from from which value
of the limits are different though this means
00:35:15.990 --> 00:35:30.170
a path dependent then this implies this limit
00:35:30.170 --> 00:35:39.950
does not exist
ok . So, hence we can easily show get whether
00:35:39.950 --> 00:35:46.190
a limit exist or it does not exist to show
the existence we have to go only through delta
00:35:46.190 --> 00:35:51.640
epsilon definition to show that the limit
does not exist we have to we have to show
00:35:51.640 --> 00:35:56.289
that from 2 different paths values of the
limit are different ok so.
00:35:56.289 --> 00:35:57.979
Thank you very much .