Sorry for posting so many of these! I'm finding them a little tricky and keep running into problemsProve that is continuous at the origin.
I've done two proofs for this, one uses and , the other uses sequences.
The epsilon-delta version:
I need to find when .
Since i'm trying to find continuity at 0, let .
Therefore I need to find when .
For , . Hence pick .
For , . Hence pick
Therefore choose
That was preliminary work so I have to write out the proper proof:
Let and define . When :
for .
for .
Hence f is continuous .
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Now for the limits version.
Let and .
Clearly and .
I need to show that if exists.
Therefore so exists.
Hence the function is continuous at the origin.
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Thanks in advance to anyone who posts!
umm, I don't suppose you could show me what you mean?
I thought the point was that there are two sequences that tend to the same limit- one from above and one from below. Then if you can show that when the function is applied to the limit of these sequences they produce the same value. This shows that a right hand limit exists, a left hand limit also exists and therefore the function is sequentially continuous.
Showcase.
The above proof is 100% correct,δ is indeed the minimum of { ε, }.
To write your proof in a better style:
let ε>0 , then choose δ= min{ ε, } which implies .............................................1
.................................................. ............2
Now let xεDf ( Df = domain of f) and |x|<δ (and not 0<|x|<δ as you write in your proof,because you want x to take the zero value).
And now we have two cases:
Case 1) For, xεDf and x<0 and |x|<δ ====> (by using (2))
Case 2) For,xεDf and and |x|<δ =====>( by using (1)) .
Since for all x,belonging to the real Nos
But ,
Hence xεDf and |x|<δ =====> |f(x)|=|f(x)-0|=|f(x)-f(0)|<ε.
Thus f is continuous at x=0
How can you explain that phenomenon that happens very often in mathematics and particularly in analysis,where one writes a proof and then he/she cannot check his/her own proof and be sure of its correctness???
Thanks! I had never come across a situation where was the minimum of two functions of and that made me a little confused =S
Do you also think my limit proof is a little iffy?
Here's my second attempt:
Let be any sequence such thatProve that is continuous at the origin.
Therefore and .
As before, I need to show that if exists.
Hence:
if .
if .
The limits are the same so the function is continuous at the origin.
Is this method better?
In your preliminaries in the ε-δ proof you wrote and i quote:
"For x>0, .Hence pick δ=ε.
For x<0, .Hence pick "
Here you put δ=ε and ,instead of putting and .
And so when you wrote let δ= min{ },it was = min{ }.
So you have come across a situation where δ is a function of ε .Only that the function of ε is hidding behind the .
In your ε-δ proof what you actualy did is that:
You used an example to justify the general theorem:
,instead of using that general theorem to prove your example.
Now in proving that general theorem ,in the converse proof:
,we come across two deltas
For the sequence part i am still working on it
hmm, I see that I did the first limits problem incorrectly since I proved it for only one set of sequences, not for all sets.
What was wrong with my second attempt? I'm can't see anything wrong with it since I proved the function as continuous at the origin for all sequences that tend to zero.
Again your proof is correct except perhaps a few typo mistakes .
To write it again in a better style ,here it is:
Let { } be a sequence in interval [ 0,+ ) ,for , and .Then ,and { } is in [ 0,+ ).
Where for all n belonging to N.
But { } belongs to the (- ) interval,for x<0, and .
Now since the functions ,x^2, and sin(x) are continuous over the real Nos R and thus at x=0 we have that:
,and .
Hence f is continuous at x= 0.
Of course there are a lot of theorems involved in the above proof and i think it would be a very good exercise if you could check thru the theorems the steps of your proof .I did that myself ,but you know the risk of making a mistake is very easy.
And of course it would be easier if you used two different general sequences { } ,{ } belonging to the intervals [0,+00) and (-00, 0) respectively instead of using the same sequence and running to the risk of making unforeseen mistakes.
I initially tried doing it just like this but when I took the modulus of both and they became defined on the same interval. This caused a huge problem because I didn't have a right and left hand limit, only a right hand one.And of course it would be easier if you used two different general sequences { } ,{ } belonging to the intervals [0,+00) and (-00, 0) respectively instead of using the same sequence and running to the risk of making unforeseen mistakes.
Is the method practically the same but instead of we have ?
You do not have to take the modulus of either sequence.
All you have to do is:
Let { } be in the (- ,0) and .
Also let { } be in [0,+ ) and .
Now again the functions ,x^2, sin(x) are continuous we have:
,and
Hence f is continuous at x=0 .
Note ,since { } is in (- ,0) then is a limit from the left of ,0.
For the same reason( { } is in [0,+ )) , is a limit from the right of ,0