How can I prove that $\lim\limits_{n\to\infty}\frac{n+\sin n}{n+1} = 1$

Question

How to prove $\displaystyle \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1$ ?

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

Note that $$\left|\frac{n+\sin(n)}{n+1}-1\right|=\left|\frac{-1+\sin(n)}{n+1}\right|\leq \frac{1+|\sin(n)|}{n+1}.$$ Now recall that $\sin(n)\in [-1,1]$. Can you take it from here?

Answer 2

Start out by separating your numerator: $$\lim{\frac{n+\sin(n)}{n+1}}=\lim{\frac{n}{n+1}}+\lim{\frac{\sin(n)}{n+1}}$$ The first term goes to one, and the second to zero, giving a final answer of one.

Answer 3

First, devide the numerator and denominator by $n$

$$ \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} =\lim\limits_{n\to\infty}\frac{1+\sin(n)/n}{1+1/n}=1$$

and then use $\lim_{n\to \infty}\frac{\sin(n)}{n}=0$ as $|\sin(n)|\leq 1$.

Answer 4

$$\dfrac{n+\sin(n)}{n+1}=\dfrac{1+\frac{\sin(n)}{n}}{1+\frac{1}{n}}$$

Answer 5

You need to show that for $ \epsilon \gt 0$, there is an $N_0(\epsilon)$ such that for $n\gt N_0$:

$|\dfrac{n+\sin(n)}{n+1}-1| \lt \epsilon.$

Let $ \epsilon \gt 0 $ given.

Choose: $N_0$ such that $\dfrac{2}{\epsilon} \lt N_0 .$

For $n \gt N_0:$

$\bigl| \dfrac{n+\sin(n)}{n+1} -1\bigr|$=

$|\dfrac{\sin(n) -1}{n+1}| \lt |\dfrac{2}{n}| \lt \epsilon$.