Prove the limit using the Epsilon-Delta definition?

Question

My attempt:

Let ϵ > 0 be given and define N = -$\frac{ϵ+1}{ϵ}$

It follows that for all n>N,

n > -$\frac{ϵ+1}{ϵ}$ $\iff$ $\frac{1}{n}$ < -$\frac{ϵ}{ϵ+1}$ $\iff$ $\frac{1}{n}$ + 1 < -$\frac{ϵ}{ϵ+1}$+1 = $\frac{1}{ϵ+1}$ $\iff$ $\frac{1}{1/n + 1}$ > ϵ+1 $\iff$ $\frac{n}{1+n}$ > ϵ+1 $\iff$ $\frac{n}{1+n}$-1 > ϵ $\iff$ $$\lim_{n\to\infty} a_n = 1 $$

Answer 1

Hint: Remember that $N \in \mathbb N$. Notice that

$$\left|\frac{n}{n+1} - 1 \right|= \left|-\frac{1}{n+1}\right| = \left|\frac{1}{n+1}\right|$$

There exists $N \in \mathbb N$ such that $\frac{1}{N+1} < \epsilon$.

Answer 2

Scratch paper: Choose $\epsilon>0$. Suppose the claim is true, then there should be a natural number $N$ such thta whenever $n\ge N$, $|\frac{n}{n+1}-1|<\epsilon.$ Which is equivalent to $1-\frac{n}{n+1}<\epsilon$. So, $1-\epsilon<\frac{n}{n+1}$. So $n$ must satify that $n>\frac{1-\epsilon}{\epsilon}$.Take $n$ to be the smallest integer satisying that.