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the teacher solved an exercise for example, and I don't understand why the solution is right. I have been trying to understand the solution for hours. unfortuently I still don't understand.

proof that: $$\lim\limits_{ n \to \infty}\frac{1}{n}=0$$

solution:

we need to proof that, for every epsilon, there is a N that for every an > N :

$$| \frac{1}{n}-0| < \varepsilon$$ ( I understood that in order to proof that 0 is the limit, I have to proof that the limit definition is true about this exercise )

$ \frac{1}{n} $ is always positive, because n is always positive, so: $ | \frac{1}{n} | = \frac{1}{n} $

$ \frac{1}{n} < \varepsilon $

$ n > \frac{1}{\varepsilon} $ ( this part I didn't understand. what does it mean n > expression ? why do I have to get to " n > expression" in every lim proof? [that's what our teacher said])

The teacher wrote this example, here I understood nothing:

$ \varepsilon =\frac{1}{1000} $ so n > 1000

thus,$ N = \frac{1}{\varepsilon} $

please help me

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Essentially, by the limit you (try to) capture all numbers bigger than $n$ by an arbitrarily small $\varepsilon$. The proof is there to show that your capturing effort (by given $\varepsilon$) really works for all numbers bigger than $n$.

By showing this, you can show that you can drive the $\varepsilon$ to any imaginable proximity to $0$ even to such distance, that $\varepsilon$ and $0$ become indistiguishable i.e. that the limit is equal to $0$ and still there will be an $n$ for which the above capture works up to infinity.

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  • $\begingroup$ if I get n > expression, does it mean that the limit I supposed to proove is right? for example ,what result would I get if the limit in the exercise isn't the right limit? $\endgroup$ – Silas2033 Nov 18 '15 at 7:10
  • $\begingroup$ If your computation happened to show an $n$ for which the inequality does not hold, you would have proven that the limit does not work. $\endgroup$ – Thinkeye Nov 18 '15 at 8:00
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Let $x_n=1/n$ then $$\left|x_n-0\right|=\frac{1}{n}<\varepsilon\tag{*}$$ for any $\varepsilon>0$. We need to find a suitable $n$ that this holds for any arbitrarily small $\varepsilon$ and we get it via $$\frac{1}{n}<\varepsilon\leftrightarrow n>\frac{1}{\varepsilon}.$$

This basically means that you can give me any accuracy $\varepsilon$ and I can find a $n$ s.t. $x_n<\varepsilon$. This is true for all $\varepsilon>0$. For example you pick $\varepsilon=0.5$ then I can see that I have to chose a $n>1/0.5=2$ which means any $n\geq 3$ would satisfy the equation $(*)$.

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  • $\begingroup$ so the "n" in the inequality , is "N" in the lim definition? $\endgroup$ – Silas2033 Nov 18 '15 at 7:06
  • $\begingroup$ @Silas2033 For the proof we use the following definition: Let $(x_n)_n$ be a sequence then $a\in\mathbb{R}$ is the limit of $(x_n)_n$ if there is a $N\in\mathbb{N}$ for each $\varepsilon > 0$ s.t. $$|x_n-a|<\varepsilon\qquad \text{for each }n\in\mathbb{N}\text{ with }n\geq N.$$ In my example there is such an $N$ for $\varepsilon=0.5$ with $N=3$ because the inequality holds for each $n\geq N$. $\endgroup$ – Christian Ivicevic Nov 18 '15 at 9:37
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What you need to understand first is that $\varepsilon $ is a fixed positive number which is intuitively assumed to be very small. So by proving the statement for $\varepsilon$ without giving any initial value to it, you in fact prove it for any positive number that is fixed in advance. This leads to the inequality: $ n > \frac{1}{\varepsilon} $. For better understanding, take an initial value like your teacher did, $\varepsilon=1/1000 $. This will lead you $ n > \frac{1}{1/1000} $, i.e. $ n > 1000 $. From here take as $N=1000$ beacause for any $n>N=1000$ the statement will be true for the given $\varepsilon=1/1000$

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