Prove by using definition of limit

[furi0n]

lim f(x)x->0= does not exist

fx={1 if x is rational,
0 if x is not rational} by using definition of limit


My teacher had asked us to this question,but next lesson She does not explain how to solve this, maybe she will ask us to it in exam. but i know why it does not exist but i can't explain help please...

 

[arildno]

Consider the two subsequences of x's converging to zero, namely 1) rational x's 2) irrational x's

What is the limit of f at x=0 when evaluated along 1)?
And what is the similar limit of f when evaluated along 2)?

What can you conclude from this?

 

[HallsofIvy]

Suppose there is a limit, say "a".
The definition of limit says, given any \epsilon> 0, there exist \delta> 0 such that if |x|< \delta then |f(x)- a|< \epsilon.

Take \epsilon= |a| or \epsilon= |1- a|, which ever is smaller, and consider what happens, in |f(x)- a|, if x is rational and what happens if x is irrational.

 

[furi0n]

Suppose there is a limit, say "a".
The definition of limit says, given any \epsilon> 0, there exist \delta> 0 such that if |x|< \delta then |f(x)- a|< \epsilon Take \epsilon= |a| ...

if i take \epsilon= |a| then a- |a|< f(x) what i understand is that we try to make 0 left side of inequality if we do, we should tkae epsilon a but then epsilon may be negative or we should suppose limF(x)=|a| is this correct?

 

[HallsofIvy]

I don't understand what you mean by "epsilon may be negative" when you hve defined \epsilon= |a|. Nor do I know what you mean by "try to make 0 left side of inequality".

 

[furi0n]

sorry about this meaningless sentence now i don't remember what i try to said. :D
"consider what happens, in |f(x)- a|, if x is rational and what happens if x is irrational"
if x is rational then fx will be 1 then |1-a|<epsilon if i choose epsilon= |1-a|/2 then |1-a|>epsilon this is false. if f is irrational then fx will be 0 then |-a|<epsilon if i choose |a|/2 then epsilon<|-a| then it will be false ,too. what i did all is correct, isn't it ? please can you say only my mistake i want to understand how i should think when solving these question. :D

 

[HallsofIvy]

You can't "choose epsilon= |1-a|/2". Epsilon is given and you have no control over it. I said before 'Suppose there is a limit, say "a".' Either a\le 1/2 or a&gt; 1/2. If a\le 1/2 and \epsilon= 1/4 or less, do you see where "f(x)= 1 if x is rational" will cause a problem? If a&gt; 1/2 and \epsilon= 1/4 or less, do you see where "f(x)= 0 if x is not rational" will cause a problem?

 

[furi0n]

yes i see but how do you found epsilon 1/4 yu said me you can't choose but now you said epsilon=1/4 i understand rest of them.. do you know any link which i can learn exactly how to prove such problem..

 

[HallsofIvy]

Please be more careful in reading what I write! I did not "find" epsilon equal to 1/4. I said "IF" epsilon is 1/4 or less, then...

Because neither you nor I can choose epsilon, "|f(x)- a|&lt; \epsilon" must be true for all epsilon. I just pointed out that it is NOT true for some values of epsilon.

 

[Landau]

furi0n: it looks like you have problems with the very definition of 'limit'. Have you ever worked with this definition before, in simple cases? E.g. can you prove that if f(x)=x, then the limit of f(x) as x->a is a?

 

[furi0n]

of course i have studied but i have problem unfortunatly about details. some special problem is more hard than other i could solve.i looked different books i understood but i can't solve some problem still. :d i found the solution. i attached solution if you look solutıon ou can see "given epsilon=1/2" then what does it mean?.

 

[dimitri151]

See the attached pdf.
You don't get to pick epsilon in the sense that for f to have limit L the limit statement must hold for all epsilon >0. In the case of that other solution the epsilons are picked to disprove that there's a limit.