Proving $\lim_{x \to a} \frac{x}{1+x} = \frac{a}{1+a}$ I am trying to prove that, for all $a \in \mathbb{R}, a \neq -1$,  
$$\lim_{x \to a}  \frac{x}{1+x} = \frac{a}{1+a}$$
by using the $\epsilon - \delta$ formalism. I'm completely lost and unsure on how to start.
 A: First we observe that, if $\lim_{x \to a}\frac{x}{1+x} = \frac{a}{1+a}$ or $\lim_{\xi \to 1+a}\frac{\xi - 1}{\xi} = \frac{a}{1+a}$, then 
$$
\lim_{x \to a}\frac{x}{1+x} = \lim_{\xi \to 1+a}\frac{\xi -1}{\xi};
$$
hence it suffices to prove for the map $\xi \to \frac{\xi -1}{\xi}$ with $a \neq -1$.
If $\xi \neq 0$, then
$$
\bigg| \frac{\xi -1}{\xi} - \frac{a}{1+a} \bigg| = \frac{|\xi - (1+a)|}{|\xi||1+a|};
$$
if in addition $|\xi - (1+a)| < \frac{|1+a|}{2} =: \delta_{1}$,
then by triangle inequality we have $||\xi| - 2\delta_{1}| \leq |\xi - (1+a)| < \delta_{1}$,
implying that
$\delta_{1} < |\xi|$, implying that
$$
\frac{|\xi - (1+a)|}{|\xi||1+a|} < \frac{|\xi - (1+a)|}{2\delta_{1}^{2}} =: M_{\xi};
$$
given any $\varepsilon > 0$, we have $M_{\xi} < \varepsilon$ if in addition $|\xi - (1+a)| < 2\delta_{1}^{2}\varepsilon =: \delta_{2}$.
All in all we conclude that, for every $\varepsilon > 0$, for $\xi \neq 0$ and $|\xi - (1+a)| < \min \{ \delta_{1}, \delta_{2} \}$ we have
$$
\lim_{\xi \to 1+a}\frac{\xi-1}{\xi} = \frac{a}{1+a}.
$$
