# Prove that $\lim_{n\to\infty} \frac{1+\sin{(n)}}{n} = 0$.

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?

Hint: Use that $$\left|\frac{1}{n}+\frac{\sin(n)}{n}\right|\le \frac{2}{n}$$ since we have $$\left|\sin(n)\right|\le 1$$
$$|\frac {1+\sin\, n} n| \leq \frac 2 n<\epsilon$$ if $$n >\frac 2 {\epsilon}$$.
The previous answers are great. In a more general case, if $$f$$ is a bounded function, you may want to simplify your expressions by replacing its actual value for the bound, so, knowing $$\forall x \in \mathbb{R}, (1+\sin(x)) \in (0,2)$$:
$$|\frac{1+\sin{x}}{n}| \leq \frac{2}{n} \to 0$$