# Show limit does not exist for a =/= 0

1. Dec 22, 2008

### snipez90

In another thread, I explained that it is easy to show that the function defined by f(x) = x, if x is rational; 0, if x is irrational, approaches 0 at 0 using epsilon-delta. However, I am stuck in trying to show that f does not approach any number at a if a =/= 0 using the epsilon-delta definition for limits (NOT the sequential definition, which I will learn soon).

Assuming the fact that the rationals and irrationals are dense in the reals, it's easy to show that f does not approach any number near a for the function f(x) = 1, if x is rational; 0 if x is irrational. But when x is allowed to vary, I'm not sure which epsilon I should be choosing. I always end up with two sets of inequalities that do not seem to lead to a contradiction. Any hints are appreciated.

2. Dec 22, 2008

Negate the definition of the limit. The statement that the limit of f(x) as x goes to a does not exist is this: for all values L, there exists $\epsilon > 0$ such that for all $\delta > 0$, there is some value x such that $\lvert x - a \rvert < \delta$ and $\lvert f(x) - L \rvert \ge \epsilon$.

I think a choice of $\epsilon = \lvert a/2 \rvert$ should work for any nonzero a. Depending on L, you'll have to pick x to be either a rational number or an irrational number.

3. Dec 22, 2008

### sutupidmath

Using the def of the limit in terms of the sequences it is not difficult to show that the limit of this function does not exist, since we can find two subsequences that converge at zero, but their corresponding sequences of the function converge at different points.

I am not sure whether the following would be considered a totally correct way of showing that the limit of this function does not exist using epsilon delta definition of the limit, however i will give it a try.

Let's suppose that the limit of that function exists, that is

$$\lim_{x\rightarrow a}f(x)=L$$ where a is different from zero.

Now, i broke the problem into 3 subcases.

1. Lets suppose that L>0

Then according to a theorem there exists a $$\delta$$ neighbourhood of a such that for every x in $$(a-\delta,a+\delta)$$ we will have f(x)>0.

However this is not possible since from the Completness axiom in the interval $$(a-\delta,a+\delta)$$ we will have irrational numbers, so from the way the function is defined we would have 0>0 which is a contradiction.

2. Let's suppose that L<0

Now as above, there exists a $$\delta$$ neighbourhood of a such that whenever x is in $$(a-\delta,a+\delta)$$ we will have f(x)<0. But again since we will have irrational numbers there, from the density of reals it would mean that 0<0, which is not possible.

3. Let's suppose that L=0.

Now from the definition of the limit we would have that

$$\forall \epsilon>0,\exists \delta>0$$ such that , whenever

$$0<|x-a|<\delta$$

$$|f(x)|<\epsilon$$

But from $$0<|x-a|<\delta$$ we have

$$a-\delta<x<a+\delta$$

Now if we choose $$\epsilon=a-\delta$$

when x is rational we will have at the same time

$$-\epsilon<x<\epsilon$$ and

$$\epsilon<x$$ which is not possible.

SO, i guess, these would establish that such a limit does not exist.

Edit:To adriank. What do you thin of this proof, would this work as well?

4. Dec 22, 2008

### lurflurf

yuck
lim sup |f|=|x|
lim inf |f|=|0|=0
in order that the limit exist we require
|x|=0
clearly this is true if and only if
x=0

5. Dec 22, 2008

sutupidmath: I only briefly read your post (so some details might have been missed), but it looks perfectly good. My idea when I posted was fixing epsilon to be a/2, if |L| < |a/2|, then you pick x to be a rational number near a, and otherwise, pick x to be a rational number near a. I think I like yours better, though. (I think I had this very question on an analysis assignment from last year; I can't remember what exactly my solution was. I'll have a look.)

lurflurf: The original poster specifically was asking about using the epsilon-delta definition of the limit.

6. Dec 22, 2008

### lurflurf

I did use epsilon-delta definition of the limit. My epsilon's and delta's are shy, but perhaps some would like to see them.
let
f:R->R
with
f(x)=xQ(x)
where Q is the charactistic function of the rational numbers.
we observe that
0<=|f|<=|x|
since lim f=L->lim|f|=|L|
we will find it helpful to work with |f| initially

We say a function f has a lim sup LS as x approches a if for any epsilon>0 there exist delta>0 such that
f(x)<L+epsilon whenever 0<|x-a|<delta, and there exist x with 0<|x-a|<delta such that LS-epsilon<f(x).
also
We say a function f has a lim inf LI as x approches a if for any epsilon>0 there exist delta>0 such that
L-epsilon<f(x) whenever 0<|x-a|<delta, and there exist x with 0<|x-a|<delta such that f(x)<LS+epsilon.

now show (hint every real number is near rational number and an irrational number (In fact an infinite number of each.).)
LS=|a|
LI=|0|=0

We say a function f has a lim L if LS<=L<=LI.
Thus if |f| has a limit L=0 and a=0
The original poster has shown the zero case.
We may conclude f has a limit if and only if a=0, in which case the limit is 0.

For others that have read up to the subsequence part I offer an alternative. Take limits over rationals and irrationals. The limit exist only when all sublimits exist and are equal.

7. Dec 22, 2008

### lurflurf

The whole point is to chose any epsilon, perhaps you ment which delta?
Consider a=e~2.718281828...
for any choice of delta we will have some rational and some irational x.
let u be irrational with 0<|v-e|<delta
|f(v)-L|=|-L|<epsilon
This will only be true for all epsilon if L=0.

8. Dec 22, 2008

### snipez90

Wow, thanks for the replies. I think my biggest failure was overlooking the idea that you can consider different cases for L. In the examples I have encountered, I just left L alone and showed a clear contradiction based on the width of the interval on the y-axis (take sin(1/x) or |x|/x near 0 for example).

adriank: I think your proof might be the simplest in terms of not having to use any theorems or definitions besides the e-d definition. However, do we have to absolute value the L? Whether we consider |L| or L, the case |L| or L $$\geq$$ |a/2| is easy, but how do you prove the case for L or |L| < |a/2|? Also, I find it somewhat more congenial to assume the limit does exist and then show contradictions but this is probably an inflexibility on my part.

sutupidmath: Great proof! The only theorem you seem to use is the sign theorem that's usually associated with continuity but in this case would work well since L is broken down into cases. Did you know immediately know to separate the cases?

lurflurf: I'll have to look at the definitions for limsup and liminf since I've only heard of them. As for your last post, I meant the choice of epsilon in the negation of the epsilon-delta definition. However, I think I follow your argument that if the limit exists, it can ONLY occur at a = 0. Basically, you are considering the f(x) = 0, if x is irrational part. Since the irrationals are dense, this means that if the limit does exist, then there is some x in (a - d, a + d) s.t. |0 -L| < e which implies |L| < e. But this last inequality has to hold for all epsilon so L is forced to be 0. Please correct me if I'm wrong; I think the argument is nice.

Anyways this example (not chapter problem) was the one I could not prove in Spivak's chapter on limits. He basically stated, "By now you should have no difficulty convincing yourself that it is true".

9. Dec 22, 2008

For the case of |L| < |a/2|, the idea is that the epsilon-neighborhood of L doesn't intersect the line y = x, so for any delta, you can pick a rational number x such that f(x) doesn't lie in the epsilon-neighborhood of L.

The reason I used absolute values was just so I wouldn't have to deal with the cases a > 0 and a < 0 separately.

I did find my solution for this question on one of my assignments, but it used the sequential criterion, so that doesn't help.

10. Dec 22, 2008

### lurflurf

It can be important to consider different L, such as when you define a function with a limit (ie let f=lim g) or when (like now) you want to know for which a does the limit exist. I think you understand well my point, since f=x for rationals and f=0 for irrarionals a limit can only exist when x=0. The lim sup and lim inf together are just a different way of evpressing the same idea as a limit. Apart they represent to parts of a limit. The idea of a limit is we trap f in (L-epsilon,L+epsilon). The idea of a lim sup is we trap the top of f in (LS-epsilon,LS+epsilon). The idea of a lim inf is we trap the bottom of f in (LI-epsilon,LI+epsilon). If LS and LI agree we can trap all of f, if they do not we cannot trap f.

11. Dec 23, 2008

### snipez90

Hmm I just have one more question. Why could sutupidmath choose a - d to be epsilon? What if a < 0, then certainly a - d < 0. But then we can't have that as epsilon. Would we choose e = -(a-d) = d-a?

12. Dec 23, 2008