I'm trying to solve the problem of showing that $$\forall x \neq -1$$ $$\lim_{x\to a}\left(\frac{x}{1+x}\right) = \frac{a}{1+a}$$ using the $\epsilon$-$\delta$ definition of a limit.
I've broken it down till $$\frac{|x-a|}{|(1+x)(1+a)|}$$
I now need a lower bound for $$|1+x|$$
I realize this is probably really simple but I'm just scared of making a mistake.
Given that a can be either positive or negative I'm not sure how to pick the lower bound exactly?
Note $a\neq -1$. So $|a+1|>0$. If we select $\delta<\frac{|a-(-1)|}{2}$ ( of course with other conditions related with $\epsilon$), then for $|x-a|<\delta$, we should have $$|x+1|\geq |a+1|-|a-x|>|a+1|-\frac{|a-(-1)|}{2}=\frac{|a+1|}{2}>0$$
This gives you a lower bound. The idea is that we force $x$ be closer to $a$, so the distance from $x$ and $-1$ must be larger than half of the distance between $a$ and $-1$.