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!
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: $$\lim_{n \to \infty} a_n= 2$$