00:01
So the first function is, so the first function is fx is equal to x upon x square, x upon x squared minus x.
00:14
So this is the first function.
00:16
So i can write this as x upon x into x minus 1.
00:21
The denominator should not be 0.
00:23
So out of domain, the function fx is not defined at x is equal to 0 and x is equal to 1.
00:31
The function is not defined at this point because we cannot make the domain to be 0.
00:38
Exact 0.
00:39
It can't be exact 0.
00:41
But so our point of consideration would be x is equal to 0 and x equal to 1.
00:47
First of all, let's check the left hand limit of, that is limit x tends to 1 minus limit x tends to 1 minus x upon x into x minus 1 minus x upon x into x minus 1 minus 1.
01:01
Which is equal to so it would be one upon one and one minus two just so this will this will be tending to this will be very zero but a negative zero so this will be tends to minus infinite and the right hand limit will be limit x tends to one plus so it would be x upon x into x minus one so it will tends to plus infinite so since left -hand limit is not equal to right hand limit, left hand limit is not equal to right hand limit.
01:39
Therefore limit does not exist.
01:41
Therefore limit at this point does not exist and it is discontinuous, discontinuous at x tends to 1 and the discontinuity is infinite discontinued...