# Epsilon-Delta Proof Lower Bound help!

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?

• Can't you just use the fact that both limits exists and the lower limit is never zero, therefore the limit of the divison of the functions equals the division of the limits? – Rono Nov 4 '14 at 3:33
• Specifically says to use epsilon delta, I cant use the Limit Laws. – Exc Nov 4 '14 at 3:39

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 $-1$.