Example continuity at point - Differentiability Video Lecture - Class 12

hello friends this video on continuity
and differentiability part four is
brought to you by exam feel calm no more
fear from exam before watching this
video please make sure that you have
watched part one to power three so one
thing is clear is the continuity is
defined at a point correct I told the
continuity and discontinuity is
different at the point so let us take
some examples to clarify our concept so
the question says we have a graph where
X equal to 3 for X is less than 0 less
than equal to 0 and X is equal to 4 for
X is greater than 0 so we have this
graph X is less than equal to 0 we have
value 1 to 3 so we have the graph like
this this guy is 3 so this point is 4 0
please note it is less than equal to 0
and this is greater than 0 so for less
than equal to 0 that means equal to 0
also part of this so we take from here I
don't take this fine greater than equal
to 0 so that means at 0 this guy is not
parts well create a hollow hole here and
his solid hole so hollow hole means this
is not included so X is going to 0
I'll see if I is equal to 3 not 4
correct but it is hollow that means you
have to exclude this but it is solid
that means you have to include this this
is my graph actually for this now we
have to find whether it is continuous or
not it this guy X is good to 0 because
you can see just by looking at the graph
I can tell that it is discontinuous
function why because to create this
graph I have to lift my pen just but if
you have if you have the graph draw the
graph and by looking at the graph you
can tell where the function is
continuous or discontinuous but
mathematically also if you want begin
through we can find the limit I'll show
you how
the rules is that limit of effects
extends to see here sees are zero
because we are talking about the
continuity at point zero because at zero
we had I see this should be equal to F
of zero by because C is placed by zero
so the first thing is let's find the
limit exists or not for this guy so left
hand limit is nothing but the limit X
tends to 0 minus or effects there is
nothing but number very very less than
zero that does you can say nothing but f
of minus 0.0001 correct you guys left
hand limit I'm approaching zero from the
left hand side so number that is not
zero but just on the left hand side of
zero that is point zero zero one for
this value for this value what I get
this is nothing but three so my left
hand limit is three now let me find the
right hand limit this is nothing but
limit X tends to 0 plus f of X so I am
approaching this from the right hand
side so I'd want number that is not zero
but a number on the right hand side of
zero just just greater than zero so I
can take this guy's nothing but f of
0.00001 a number that is just greater
than zero and if you take such point do
you it Y is 4 so f of this is 4 so my
left hand limit this is not equal to
right hand limit so in this case the
limit itself doesn't exist the limit
doesn't exist the next question was to
check that limit value is equal to f of
0 but that is not required now because
the limit itself does not exist you can
see that
with discontinuation correct in caseware
limit exist the second step is we have
to find the value of f of 0 and then we
have to check because our is equal so we
are done I think in the last slide where
the limit existed but the number the
value was not equal to f of C so there
also it was discontinuous we'll take one
more example here if you see the example
is like this for X is not equal to 0 it
is 3 1 2 3 & 4 X equal to 0 it is 4 so 4
X going to 0 the value is 4 and for
other it is 2 so it is this kind of
graph correct here also looking at the
graph we can tell it is discontinuous
because to draw the graph I have to let
my pen but mathematically you have to
find so mathematically we have to just
find if the limit of X tends to C F of C
is equal to F of C here C is equal to 0
because the point of contention is 0 and
point at 0 X equal to 0 we have to find
whether it is continuous or not so what
we can do let's find first left-hand
limit is nothing but limit X tends to 0
negative f of X is I'm approaching the
value of x from left hand side that is
any number there is not 0 but just
smaller than 0 so I can take this guy f
of minus 0.0001 so if you take this guy
what you get the value 3 so you find the
value of F of minus point zero zero the
one you get the get the value of 3 now
let's find right hand limit that is
nothing but limit X tends to 0 plus f of
X approaching the guy from right hand
side
any number that is greater than zero is
just greater than zero but not equal to
zero that is I again take that value as
plus point zero zero zero zero one any
number I can take just greater than zero
see if I take that number that kind of
number I get Y as again three so I what
I have seen here left hand limit is
equal to a right angle
yup P my left hand limit is equal to
right hand that means the limit exists
so I can say that limit of X tends to 0
f of 0 has a value and that value is
equal to 2 now my first step is clear
that means limit existed the second step
is I have to check whether that guy
value is equal to F of 0 because here C
is equal to so let's find F of 0 F of 0
is nothing but 4 so what I can see F of
0 is 4 and the limit of f of 0 X tends
to 0 therefore X breaks
there's nothing but 3 this is f of X
please no so I can see that both are not
equal FX extends to 0 limit is equal to
3 and F of 0 is well they are not equal
since they are not equal I can say that
they are not continuous here also you
can say that 0 is point of discontinuity
0 is the point of discontinuity so what
I've done in this case I saw that the
limit existed but that value was not
equal to f of C so with this for
examples I think you can for a given
graph and for a given point
please note for a given graph and for a
given point you can find that whether
that particular graph is continuous or
discontinuous at a particular point
because both examples we took was we had
a graph we had a point and we wanted to
find whether the
is continuous or discontinuous let us
take one more example here we have to
clue that the function FX 5x minus 3 is
continuous at X is equal to 0 and X is
equal to 5 so they have to approach
first we can draw the graph and then we
can show that at X is equal to zero and
X is equal to 5 the function is
continuous that means we don't need to
lift the pen or the next approach is we
know that if we have FC right or if X
limits extends to C this is equal to FC
then I can see that the function is
continuous at Point C so sheet of C will
put 0 and 5 and it will try to very bad
way so first let's see if the function
is or continuous at X is equal to 0 so
for this part it will do first we'll
check if limit exists or not so what we
can do is we'll find left-hand limit of
left hand limit at F is it so that is
nothing but limit X tends to 0 minus f
of X so you can take any value which is
just less than 0 against a minus 0.00001
correct this value I can take so this
will become almost 5 into minus 0.0001
- this way the almost zero so that
becomes almost -3 so the left-hand limit
for this function is minus 3 similarly
let's find the right-hand limit at F is
equal to 0 so that will be nothing bad
never mind FX near X tends to 0 plus
here please note it is minus here this
time this plastic so let us take any
value which is just greater than 0 so I
can take 0.000001 this is nothing but 5
into zero point zero zero zero zero one
- tea there is almost zero she does
becomes almost - so what we have seen
the left hand limit and the right hand
limit are equal that means the limit
exists so I can say that limit FX we are
extends to zero exist and that value is
minus three the next thing is you shake
with that this guy is equal to f of C
see here is zero that is f of zero so
what is f of 0 F of 0 is nothing but 5
into 0 minus 3 and that is per thing
well - Lee so what we have seen both
these values are equal
seeing this values are equal I can say
that limit of FX X tends to 0 is equal
to F of 0 thus I can say that the
function is continuous and X is again
easy
so with this I can see that X is equal
to 0 V 100% confirm that this functions
can - it may be discontinuous at other
points but at X is equal to 0 this guy's
similarly we can do it X is equal to Phi
I'll not do this for you completely but
as esteli the steps how to do it so you
have to first find the left hand limit
of had the visible define that does
limit X tends to 5 minus FX and then we
should fight the right hand limit so
right hand limit as X tends to 5 plus
there is a number just greater than
class 4 fix you get this valid they are
equal that means the limit exists and
then you define the value of f 5 so if
these values are all equal then you can
say that the function is continuation
and the answer will be yes because the
function will be continuous why because
it's a equation the linear equation in
terms of like so this kind of equation
will generally continues thank you visit
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