If exists then is it proper to say that
provided f(x) is continuous at x=a.
I read this somewhere and do not believe it to be authentic.
I have come across many examples which appear to support the above statement but then there are other examples which do not support this.
If it is true then is there a rigorous proof in support of this.
One example that I have is as follows:
In this case is
Another example that i have is:
In this case is
Would be really grateful if someone verifies this for me
Thanx in advance
Also i must mention that [x] denotes the greatest integer less than or equal to x
Now we know that when x is in the neighbourhood of 0.
Therefore, which means
This is what I want to verify.
Also it is known that in neighbourhood of 0 and
What should be value of
Is it same as
This is what I want to verify
you don't need to sketch a graph. choose then if you apply mean value theorem for on the interval then you get
thus, since is an even function, for any non-zero in the interval we must have: which gives us: for any in
that interval. thus but a similar argument works for
This is precisely what I wanted to know that while evaluating
we are required to first find out
and then apply rule of greatest integer function(I would want to know the reason.Is there a rigorous proof?)
Or should we not say that since the greatest integer should anyway be 0 whether we find
or we find
I am perplexed because in case of the answer for
the same result holds if we replace right with left and with i'll prove the claim for the right continuous case: let . then since is right continous at
there exists such that whenever also, since there exists such that whenever
finally we know that there exists such that whenever now let and then:
thus we must have: which proves that
as an example you can check your question: the function is right continuous at x = 0 and in some neighbourhood of x = 0. that's why you got that
result. if instead you take to be the ceiling function, which is left continuous at x = 0, then you'll get the same result but for the function