Epsilon-Delta Proof $ \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1$

Question

How To Construct a Epsilon-Delta proof $\displaystyle \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1$ ?

The beginning:

Fix $\epsilon$ > 0. Is there an $N\in \mathbb{N},$ such that $$n\ge N \implies\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon?$$

$$\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon \iff$$ $$\left|\frac{\sin(n)-1}{n+1}\right|<\epsilon \iff$$ $$\ldots$$

I got that $n > \frac{2}{\epsilon}-1$, is that right?

N=(max or min ?) {0, $\lfloor\frac{2}{\epsilon}-1\rfloor$}

I hope that the task is understandable. Thank you for answering.

Answer 1

Hint: Use the triangular inequality for the numerator and the boundedness of $\sin(n)$ by $|\sin(n)|\leq 1$:

$$\biggl|\frac{\sin(n)-1}{n+1}\biggr|\leq \frac{|\sin(n)|+1}{|n+1|}\leq \frac{1+1}{|n+1|}=\frac{2}{n+1}$$

Answer 2

Try these: $|x-y| \leq |x| + |y|$ and $|\sin(x)|\leq1$.