Prove by using definition of limit

In summary, my teacher does not explain how to solve the limit question, but she may ask us to solve it in an exam. The two subsequences of x's converging to zero, namely 1) rational x's and 2) irrational x's, have different limits. What can you conclude from this?
  • #1
furi0n
19
0
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...
 
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  • #2
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?
 
  • #3
Suppose there is a limit, say "a".
The definition of limit says, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x|< \delta[/itex] then [itex]|f(x)- a|< \epsilon[/itex].

Take [itex]\epsilon= |a|[/itex] or [itex]\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.
 
  • #4
HallsofIvy said:
Suppose there is a limit, say "a".
The definition of limit says, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x|< \delta[/itex] then [itex]|f(x)- a|< \epsilon[/itex] Take [itex]\epsilon= |a|[/itex] ...
if i take [tex]\epsilon[/tex]= |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?
 
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  • #5
I don't understand what you mean by "epsilon may be negative" when you hve defined [itex]\epsilon= |a|[/itex]. Nor do I know what you mean by "try to make 0 left side of inequality".
 
  • #6
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
 
  • #7
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 [itex]a\le 1/2[/itex] or [itex]a> 1/2[/itex]. If [itex]a\le 1/2[/itex] and [itex]\epsilon= 1/4[/itex] or less, do you see where "f(x)= 1 if x is rational" will cause a problem? If [itex]a> 1/2[/itex] and [itex]\epsilon= 1/4[/itex] or less, do you see where "f(x)= 0 if x is not rational" will cause a problem?
 
  • #8
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..
 
  • #9
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, "[itex]|f(x)- a|< \epsilon[/itex]" must be true for all epsilon. I just pointed out that it is NOT true for some values of epsilon.
 
  • #10
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?
 
  • #11
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?.
 

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  • #12
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.
 

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FAQ: Prove by using definition of limit

What is the definition of limit?

The definition of limit states that the limit of a function as x approaches a certain value, is the value that the function approaches as x gets closer and closer to that specific value.

How do you prove a limit using the definition?

To prove a limit using the definition, you must show that as x approaches the given value, the function approaches the limit value. This can be done by finding a value of delta (δ) that satisfies the definition, which is often done through algebraic manipulation and inequalities.

What is the importance of using the definition of limit?

The definition of limit is important because it is the fundamental concept that allows us to understand the behavior of functions near specific values. It also allows us to evaluate limits algebraically and determine the continuity of a function at a given point.

What are some common mistakes made when proving a limit using the definition?

Some common mistakes include not fully understanding the definition and its nuances, using incorrect algebraic manipulations, and not considering all possible cases when determining the value of delta (δ).

Can the definition of limit be used for all types of functions?

Yes, the definition of limit can be used for all types of functions, including polynomial, rational, exponential, and trigonometric functions. However, some functions may be more difficult to prove using the definition compared to others.

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