Prove that x[1/x] = 0 as x goes to infinity

[AdrianZ]

Homework Statement

Prove that lim(x->infinity):x[1/x] = 0 by epsilon-delta defintion. (WITHOUT USING THE SQUEEZE THEOREM)

The Attempt at a Solution

well, It's easy to prove that x[1/x] approaches 0 as x goes to infinity using the squeeze theorem, but the question is to prove that without using the squeeze theorem. I wrote down the epsilon-delta definition and then I tried to use this property that [x]<= n <-> x<= n but I failed to show that for any arbitrary epsilon there exists a N>0 such that x>N -> |x[1/x] - 1| <epsilon.

 

[HallsofIvy]

Are you using "[x]" to mean the floor of x? If so then the proof is trivial. For any x> 1, [1/x]= 0. Given any positive real number, $\epsilon$, if x> 1, $|f(x)- 0|< \epsilon$.

 

[estro]

Square brackets stands for floor/roof function?

If square brackets are the floor function then x[1/x]=0 for every x>1 so the delta epsilon is trivial.

In all other scenarios the limit is not 0.

 

[estro]

Oops didn't notice that HalloIvy already replied...

 

[AdrianZ]

yup. right. [x] stands for the floor function. I noticed that after I had already sent it on the forum. well, let's solve this one instead which is the next part of the same problem: lim(x[1/x]) = 1 as x approaches 0.
I wrote down: for any epsilon>0 there exists delta>0 s.t. if |x-0|<delta -> |x[1/x] - 1|<epsilon.
well, for any x in real numbers we have: 1/x <= [1/x] < 1/x - 1 => 1 <= x[1/x] < 1-x.
0 <= x[1/x] - 1 < -x & -x<x => 0 <= x[1/x] - 1 < |x| => |x[1/x] - 1| < |x|. Now does that tell us anything good?

 

[estro]

In these kind of problem you need to evaluate |f(x_0)-L| and show that it is small as you want it to be [you need to show that it is smaller then arbitrary epsilon] so:
$$|x[1/x]-1|\leq|$$think what you want to write here and don't forget that you have control over x, if x approaches 0 you can have x as small as you want... |

 

[AdrianZ]

so if I take epsilon to be >= |x| that definitely tells us that epsilon can be chosen to be as small as we want, but Don't I have to prove that for any epsilon that I choose there exists a delta that works in the definition above? I think the main problem here is not to show that the epsilon can get as small as I want, the main problem is to show that for any epsilon that I choose, I can choose delta s.t |x-0|<delta => |f(x) - L|< epsilon. correct me if I'm wrong please.

 

[estro]

so if I take epsilon to be >= |x|...

After this I stopped reading because here you miss the concept.
You CAN'T choose epsilon your proof should hold for every epsilon, you should show that for every epsilon there is an appropriate delta.

Think about this: the definition of a limit says that the limit exist at x_0 only and only if you can get as close as you want to f(x_0) now what can help you to get close to f(x_0)?

 

[estro]

When you want to prove that something hold for every number, you say: ok let epsilon be some arbitrary numbeR and then you show that "this" holds for that arbitrary number what in turn means that you proved it for every number.

Read carefully the definition of a limit and pay attention to every little detail.

 

[AdrianZ]

After this I stopped reading because here you miss the concept.
You CAN'T choose epsilon your proof should hold for every epsilon, you should show that for every epsilon there is an appropriate delta.

Think about this: the definition of a limit says that the limit exist at x_0 only and only if you can get as close as you want to f(x_0) now what can help you to get close to f(x_0)?

Actually I didn't miss the concept, my question was about the thing you said that because we have control on x we can make epsilon as small as we want which sounded absurd to me because we have no control over epsilon in the proof no matter what,we can't make it small or big but we should show that for any epsilon neighborhood of the Image of the function, there exists a delta neighborhood in the domain which is contained in the epsilon neighborhood of the range. read my post again. lol

well, I know what I want to have, I don't know how to get there and that's the real hardship in this problem. I need to isolate |x| in the right side and find an appropriate delta for that which is the hard part of the problem.

 

[estro]

You said that (you should put more details how you get this) $$|x[1/x]-x|\leq|x|$$, now because $$|x-x_0|=|x-0|=|x|< \delta$$ what can you conclude?
*Remember that you have control over delta...
At this point it is trivial, show me that you understand the concept.

 

[HallsofIvy]

Actually I didn't miss the concept, my question was about the thing you said that because we have control on x we can make epsilon as small as we want

No, he didn't say that. He said "we an make x as small as we want"- that is governed by $\delta$, not $\$.

which sounded absurd to me because we have no control over epsilon in the proof no matter what,we can't make it small or big but we should show that for any epsilon neighborhood of the Image of the function, there exists a delta neighborhood in the domain which is contained in the epsilon neighborhood of the range. read my post again. lol

well, I know what I want to have, I don't know how to get there and that's the real hardship in this problem. I need to isolate |x| in the right side and find an appropriate delta for that which is the hard part of the problem.

 

[AdrianZ]

You said that (you should put more details how you get this) $$|x[1/x]-x|\leq|x|$$, now because $$|x-x_0|=|x-0|=|x|< \delta$$ what can you conclude?
*Remember that you have control over delta...
At this point it is trivial, show me that you understand the concept.

well, let me formally prove the part that I claimed |x[1/x] - 1| < |x| because actually I guessed that and I feel bad about guessing something without having proved it lol

well, for any real x we always have: 1/x <= [1/x] < 1/x + 1.
x can't be 0 because It's not in the domain, therefore x is either positive or negative.
if x is positive, multiplying both sides by x yields: 1 <= x[1/x] < 1+x. adding -1 to both sides yields: 0 <= x[1/x] - 1 < x. (I)
if x is negative, multiplying both sides by -x yields: -1 <= -x[1/x] < -1 -x. adding +1 to both sides yields: 0 <= 1 - x[1/x] < -x (II).
Now according to the definition of abs(x) we can conclude that 0 <= |x[1/x] - 1| < |x|, since 0<=|x| is a tautology we conclude that |x[1/x] - 1|<|x|.
I still can't see why It's trivial but let me guess. |x|< delta tells us that |x[1/x] - 1| < delta. which tells us that if we choose delta small enough, then for any arbitrarily given epsilon we can always conclude that |x[1/x] - 1| < epsilon. We can always do that because the domain inherits compactedness from the real line and is continuous around 0. am I right? It's not TRIVIAL to me, but I think It makes sense if we have some intuition about the set that x belongs to.

No, he didn't say that. He said "we an make x as small as we want"- that is governed by $\delta$, not $\$.

I see. then I misunderstood it.

 

[estro]

...
I still can't see why It's trivial but let me guess. |x|< delta tells us that |x[1/x] - 1| < delta. which tells us that if we choose delta small enough, then for any arbitrarily given epsilon we can always conclude that |x[1/x] - 1| < epsilon.
...

Good, so what is the appropriate delta? [what delta should you use?]
Formulate you prove from the beginning to the end and keep attention to every word you say.

 

[AdrianZ]

Good, so what is the appropriate delta? [what delta should you use?]
Formulate you prove from the beginning to the end and keep attention to every word you say.

That's exactly my question. I don't know that part.

 

[estro]

You almost understand, if I'll tell you the answer at this point I'll ruin everything.
Think about it, this is very important concept...

 

[AdrianZ]

hmm, maybe I should factor out |x| from |x[1/x]-1| and then I try to find an upper bound for |[1/x] - 1/x| for a specific delta and then take min between that specific delta and the one that I get after defining an upper bound?

 

[estro]

I don't understand what you talking about...=)
Please think in other direction.

 

[AdrianZ]

like what direction?

 

[estro]

I'll give you an example and prove that $$\lim_{x\rightarrow 0}x=0$$

Let there be an arbitrary $$\epsilon>0.$$
I choose $$\delta=\epsilon/2$$ so for every x which satisfies this delta [$$|x-x_0|\leq\delta$$]the following happens: $$|x|\leq\epsilon/2$$
so
$$|f(x)-L|=|f(x)-0|=|f(x)|=|x|\leq |\delta| \leq |\epsilon/2|<| \epsilon |$$

And I'm done.

 

[AdrianZ]

Huh? Okay let me reword it. first I'll write |x([1/x] - 1/x)|= |x||[1/x] - 1/x|. then I'll choose delta to be in a specific neighborhood around 0 just to make it specific. then I'll try to find an upper bound for |[1/x] - 1/x| for that specific delta neighborhood. I'll need to make epsilon small, so I'll try to find a supremum for |[1/x] - 1/x| and then after diving both sides I'll have epsilon/sup of |[1/x] - 1/x| in that specific neighborhood. is it clear so far?

 

[estro]

I don't need to understand what you're writing, but you are then one who should try to understand what I'm writing.
Please:
1. Look at my example
2. Look at the limit definition.
3. Think.
4. Finish the problem.

 

[AdrianZ]

well, I didn't see your example first because you had asked me to fix my post or something like that, then you edited it. Your example is too easy to tell me anything that I can generalize it to this one lol. I can even prove that lim(ax+b) = am+b as x approaches m. That's not hard at all.
Solve this example for me instead: lim(x^3 + x^2 + 2x - 1) = 3 as x approaches 1. Can you find the appropriate delta for me in this particular example? This one would be a nice example, not that easy one.

 

[vela]

well, for any real x we always have: 1/x <= [1/x] < 1/x + 1.

You need to start over because you messed up right from the start: 1/x is not less than or equal to [1/x]. For example, for x=3/5, you have 1/x=5/3 so [1/x]=1, which clearly isn't greater than 5/3.