Interested to know if my method for carrying out a standard converge proof is ok.
Prove: $\lim \frac{n}{n+1}=1$
proof: (Will not be showing scratch work)
Let $\epsilon >0$ and choose an $N\in \mathbb{N}$ such that $N>\frac{1}{\epsilon}-1$. Now for $n \geq N$, implies that $n>\frac{1}{\epsilon}-1$ $\implies$ $\epsilon > \frac{1}{n+1}$ $\implies$ $| \frac{n}{n+1}-1|< \epsilon$. Therefore from the definition of convergence $\lim\frac{n}{n+1}=1 $.
[]