# Thread: Epsilon Delta proof, one side limit

1. ## [SOLVED]Epsilon Delta proof, one side limit

Hey,
I wanted to ask how can I proof the following statement using epsilon delta.
lim x/[x] = 1
x->3+

*[x] means the whole part of x, for example : [3.5] = 3, [-3.5] = -4, [3] = 3.

I do know how to proof normal statements using epsilon delta and so far I got to this point where I am stuck:
|x/[x] -1| < e
...
|x| < e/|1/[x] - 1/x|

After that i decided to limit the neighborhood of 3 to a radius of 1, therefore:
3 < x < 4
...
0<1/[x] - 1/x < 1/12

therefore:
|x| < 12e

Now I only have to choose an L so that if |x| < L, than |x| < 12e:
3 < x < 3 + L
But I cant find a proper L so that |x|< 12e.
I thought using L = 12e - 3, and than :
3 < x < 3 -3 +12e
|x| < 12e
But L must be larger than 0 for each e, so L = 12e -3 is no good.

Thanks a lot,
Michael.

2. ## Re: Epsilon Delta proof, one side limit

I didn't follow parts of your work, but it seemed that you were making |x| small (<12e). The problem requires looking at x-3 small (positive). If you make x small (i.e. close to 0), then of course you'll have a hard time showing that x is also close to 3.

$\displaystyle \lim_{x \to 3^+} \frac{x}{[x]} = 1$ means $\displaystyle \forall \epsilon >0, \exists \delta>0 \ni 3<x<3+\delta \Rightarrow \left| \frac{x}{[x]} - 1 \right| < \epsilon$

The first step is, given an $\displaystyle \epsilon>0$, to find a $\displaystyle \delta$ that works. The second step is to show that your discovered $\displaystyle \delta$ does in fact work.

1st) Find $\displaystyle \delta$:
As always, we'll look at the epsilon expression of the limit, and try to manipulate it into looking like $\displaystyle 3 < x < 3+something$, where that "something" will become the delta.

The first thing to note is that if $\displaystyle \delta \le 1$, then $\displaystyle 3 < x < 3+\delta$ implies $\displaystyle 3 < x < 4$, so that $\displaystyle [x] = 3$.
So, given $\displaystyle \epsilon>0$, we'll make sure that, whatever else we decide about $\displaystyle \delta$, we'll also have $\displaystyle \delta \le1$. This "gets rid of" that awkward function [x].

Want $\displaystyle \left| \frac{x}{[x]} - 1 \right| < \epsilon$, so, since $\displaystyle 3 < x < 4$, want $\displaystyle \left| \frac{x}{3} - 1 \right| < \epsilon$.

$\displaystyle \left| \frac{x}{3} - 1 \right| < \epsilon$ iff $\displaystyle 3\left| \frac{x}{3} - 1 \right| < 3\epsilon$

iff $\displaystyle \left| (3)\left(\frac{x}{3}\right) - (3)(1) \right| < 3\epsilon$ iff $\displaystyle |x- 3| < 3\epsilon$.

Now $\displaystyle |x- 3| < 3\epsilon$ is the desired form. Thus want to set $\displaystyle \delta = 3\epsilon$. However, must also include the stipulation that $\displaystyle \delta \le 1$. So set $\displaystyle \delta = min\{1, 3\epsilon\}$.

2nd) Prove that $\displaystyle \delta$ works:

Given any $\displaystyle \epsilon > 0$, let $\displaystyle \delta = min\{1, 3\epsilon\}$. Note that $\displaystyle \delta > 0$.

Then $\displaystyle 3 < x < 3+\delta$ implies $\displaystyle 3 < x < 3+3\epsilon$ implies $\displaystyle 0 < x-3 < 3\epsilon$

implies $\displaystyle 0 < \frac{x-3}{3} < \epsilon$ implies $\displaystyle 0 < \frac{x}{3} - 1 < \epsilon$

implies $\displaystyle \left| \frac{x}{3} - 1 \right| < \epsilon$.

Also, since $\displaystyle \delta \le 1$ and $\displaystyle 3 < x < 3+\delta$, have that $\displaystyle 3 < x < 4$, so $\displaystyle [x] = 3$.

Thus $\displaystyle 3 < x < 3+\delta$ implies $\displaystyle \left| \frac{x}{3} - 1 \right| < \epsilon$ implies $\displaystyle \left| \frac{x}{[x]} - 1 \right| < \epsilon$.

This proves that, given any $\displaystyle \epsilon > 0$, if you choose $\displaystyle \delta>0$ by $\displaystyle \delta = min\{1, 3\epsilon\}$, then

$\displaystyle 3 < x < 3+\delta$ implies $\displaystyle \left| \frac{x}{[x]} - 1 \right| < \epsilon$.

That has proven that $\displaystyle \lim_{x \to 3^+} \frac{x}{[x]} = 1$.

3. ## Re: Epsilon Delta proof, one side limit

Thanks a lot, you are awesome !

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