How do you prove indeterminate form using epsilon and delta?

Question

The question, for instance, is proving $$\lim_{x\to\infty}\frac{x}{x+1}=1$$

This is my answer, which is likely incorrect.

$$\forall\epsilon>0, \exists M \in \mathbb{R}$$ such that $$ x>M \Rightarrow \left|\frac{x}{x+1}-1\right|<\epsilon$$

$$\left|\frac{x}{x+1}-1\right|<\epsilon \iff \frac{1}{|x+1|}<\epsilon \iff |x+1|>\frac1\epsilon$$

I then get $x>\frac{1-\epsilon}{\epsilon}$

Picking $M=\frac{1-\epsilon}{\epsilon}$, I get $$x>M=\frac{1-\epsilon}{\epsilon}\Rightarrow x>\frac1\epsilon - 1 \Rightarrow \cdots \Rightarrow \left|\frac{x}{x+1}-1\right|<\epsilon$$

Could anybody please point out which part should I add or fix?

Answer 1

Your reasoning in general looks to me fine. But let me provide a "neater" argument for your reference:

If $x \neq -1$, then $$ \bigg| \frac{x}{x+1} - 1 \bigg| = \frac{1}{|x+1|}; $$ given any $\varepsilon > 0$, we have $\frac{1}{|x+1|} < \varepsilon$ if $x > \frac{1}{\varepsilon} - 1$.