If $(a_n)_{n\in \mathbb{N}}$ is a sequence in $\mathbb{R}$ and $L\in \mathbb{R}$, then $\lim_{n\to\infty} a_n = L$ if and only if for all $\epsilon > 0$ there exists an $N\in \mathbb{R}$ such that for all $n\in \mathbb{N}$ with $n\geq N$ it follows that $|a_n-L| < \epsilon$. Use this definition to prove that
\begin{equation*}
\lim_{n\to\infty} \frac{1+\sin{(n)}}{n} = 0
\end{equation*}
and \begin{equation*}
\lim_{n\to\infty} \frac{2n+1}{n} = 2.
\end{equation*}
The second one is simple enough: We have \begin{equation*} \begin{split} \left|\frac{2n+1}{n}-2\right| &= \left|\frac{2n+1-2n}{n}\right| \\ &= \left|\frac{1}{n}\right| \\ &= \frac{1}{n} \end{split} \end{equation*} for $n\in \mathbb{N}$. Now for $\epsilon > 0$, $\frac{1}{n} < \epsilon \Longleftrightarrow \frac{1}{\epsilon} < n$. So if we choose any number $N > \frac{1}{n}$, then we have that \begin{equation*} n > N \Longrightarrow \left|\frac{2n+1}{n}-2\right| = \frac{1}{n} < \epsilon \end{equation*} as required.
The $\sin{(n)}$ term is what is actually baffling me. Any thoughts?
