# $\epsilon - N$ proof confirmation.

These proofs seem to be my absolute worst problem. I just don't seem to get them, that being said, if this is right, I may have started to get the hang of it.

My limit and required assumptions:

$\lim \limits_{n \to \infty} \frac{2n + 10}{n} = 2, \epsilon \gt 0, N \in \mathbb{N}, |a_n - a| \lt \epsilon$

Proof: $|a_n - a| = |\frac{2n + 10 -2n}{n}| = \frac{10}{n} \lt \epsilon \rightarrow \frac{1}{n} \lt \frac{\epsilon}{10} \rightarrow n \gt \frac{10}{\epsilon}$

Let $N = [\frac{10}{\epsilon}] + 1$

$n \geq N = [\frac{10}{\epsilon}] + 1 \rightarrow \frac{10}{n} \leq \frac{10}{[\frac{10}{\epsilon}] + 1} \leq \frac{10}{[\frac{10}{\epsilon}]} = \frac{10\epsilon}{10} = \epsilon$

Hence $|a_n - a| \lt \epsilon$

Thank you for your time reading this, and I hope the layout wasn't cryptic! Thank you for any advice!

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Also additional query, when people use [] around their created epsilon term, eg my $[\frac{10}{n}]$, this is the floor function correct? –  Display Name Apr 16 '14 at 8:23

Let $\varepsilon > 0$ be given.
Select $N \in \mathbb{N}$, $N > \dfrac{10}{\varepsilon}$, such that for all $n \in \mathbb{N}$ with $n \geq N$ implies $$\left| a_n - 2\right| = \left| \dfrac{2n+10}{n} -2 \right| = \left| \dfrac{2n+10}{n} - \dfrac{2n}{n} \right|=\left| \dfrac{2n + 10 - 2n}{n}\right| = \left| \dfrac{10}{n}\right| \leq \dfrac{10}{N}<\varepsilon$$ Since $\varepsilon > 0$ was given arbitrarily, we now have: for all $\varepsilon > 0$, there exists a $N \in \mathbb{N}$, namely $N > \dfrac{10}{\varepsilon}$, such that for all $n \in \mathbb{N}$ with $n \geq N$ implies $$\left| a_n - 2 \right| < \varepsilon$$ Conclusion: $$\lim_{n \to \infty} a_n= 2$$
Could I perhaps trouble you for one more proof?, $\lim \limits_{n \to \infty} \frac{sin(n)}{n} = 0$ I tried to follow the same general layout, but I ended with $\frac{1}{\epsilon}$ which is useless as it inversed the inequality and therefore fails. –  Display Name Apr 16 '14 at 8:50
You should observe that $\left|\sin(n)\right| \leq 1$, can you handle it from here? –  Nigel Overmars Apr 16 '14 at 8:57
If I could possible bother you with one more query, and I believe that it will be the last on this thread. I have another question I am working on, and I have $\frac{log(n)}{n}=0$, I can't work out what my N should equal in this case, meaning, I can't workout what I am meant to do with $\frac{log(n)}{n} \lt \epsilon$ –  Display Name Apr 16 '14 at 9:30
You could use that $\log(n) < \sqrt{n}$ for all $n$. Hence $\left| \dfrac{\log (n)}{n} \right| < \left| \dfrac{n^{0.5}}{n} \right| = \left| \dfrac{1}{\sqrt{n}} \right| \leq \dfrac{1}{\sqrt{N}} < \varepsilon$ for all $n \geq N$. –  Nigel Overmars Apr 16 '14 at 9:47