Formally prove that $\lim(\frac{2n}{n+1})=2$ Taking Real Analysis for the first time and having trouble grasping what i need to show for the proof. I know i need for any $\epsilon > 0$, there exists a positive integer $N$, such that if $n \geq N$, then $$\biggl|\frac{2n}{n+1}-2\biggr|= \frac{2}{n+1} < \epsilon $$
I'm not sure where to go from here. Do I need to solve for $n$ on the right hand side?
 A: Notice first that you have a calculation mistake, it should be
$$\frac{2n}{n+1}-2 = \frac{2n}{n+1}-\frac{2(n+1)}{n+1} = -\frac{2}{n+1}$$
Now, this means that
$$\left|\frac{2n}{n+1}-2\right| = \frac{2}{n+1}$$
So it is left to show that $\lim_{n\rightarrow\infty}\frac{2}{n+1}=0$. You should  slightly modify the proof that $\lim_{n\rightarrow\infty}\frac{1}{n}=0$ see here
A: Hint:
Given that
$$\left |\frac{2n}{n+1}-2\right|=\left |\frac{2n-2n-2}{n+1}\right|=\frac{2}{n+1}<\epsilon$$ What inequality must $n$ satisfy for this to be true? What could you choose as $N$?
A: Given $\epsilon >0$. By the Archimedean's property there is an $N\in \mathbb{N}$ such that $\frac{2}{N}<\epsilon$.
Now let $m\geq N$. Notice that $|\frac{2m}{m+1}-2|=|\frac{2m}{m+1}-\frac{2m+2}{m+1}|=\frac{2}{m+1}<\frac{2}{m}\leq \frac{2}{N}<\epsilon$.
Hence the the limit is 2.
A: As an alternative note that
$$\lim_{n\to \infty} \frac{2n}{n+1}=\lim_{n\to \infty} \frac{2n+2-2}{n+1}=\lim_{n\to \infty} \left(2-\frac{2}{n+1}\right)=2-\lim_{n\to \infty} \left(\frac{2}{n+1}\right)$$
then it suffices to prove that $\lim_{n\to \infty} \left(\frac{2}{n+1}\right)=0$ which indeed corresponds to your step that is
$$\frac{2}{n+1} < \epsilon \implies n>\frac 2 \epsilon -1$$
A: Let $\varepsilon>0$ be given. By the Archimedean Property, we may find a positive integer $N$ so that $\varepsilon N>2$. Thus if $n \geq N$, then we have \begin{aligned}
\left|\frac{2n}{n+1}-2\right|&=\left|\frac{2n}{n+1}-\frac{2(n+1)}{n+1}\right| \\& =\frac{2}{n+1} \\& \leq \frac{2}{n}<\varepsilon .\end{aligned}
Therefore, $\lim_{n \to \infty} \frac{2n}{2n+1}=2$.
