How do we compute $\lim_{x\to c} \tfrac{x}{({1+x})}$ for $c\ne -1$ from the $\varepsilon$-$\delta$ defintion? I would like to compute $\lim_{x\to c} \tfrac{x}{({1+x})}$ where $c$ is a constant not equal to $-1$.
I started as follows:
$$\left| \frac{x}{1+x} - \frac{c}{1+c} \right| = \left| \frac{x-c}{(1+c)(1+x)}\right|$$
I know that the top can be made smaller than $\delta$, but I'm not sure what to do with the bottom because of the +1 and the fact that we don't know if either of those terms are positive or negative.  Any hints would be greatly appreciated.
 A: Ultimately you want
$$ \left| \frac{x}{1+x} - \frac{c}{1+c}\right| = \left|\frac{x-c}{(1+x)(1+c)}\right| $$ to be small.
Since $c \not= -1$ we can choose an interval centered at $c$ that does not contain $-1$. Let $r = \frac{|c+1|}{2}$. Then $-1 \notin (c-r,c+r)$, and if $x \in (c-r,c+r)$ the reverse triangle inequality gives you $$|x + 1| \ge |c+1| - |x - c| = \frac{|c+1|}{2} = r.$$
This leads to $$\left| \frac{x-c}{(1+x)(1+c)} \right| = \frac{|x-c|}{|1+x| \cdot |1+c|}\le \frac{|x-c|}{\frac r2 \cdot r} = \frac{2}{r^2} |x-c|.$$
Now work on your epsilons and deltas.
A: Let $\delta > 0$. Then, for $ c \in \mathbb R \setminus \{-1\}$ it is : $|x-c|<\delta$. Now, it is :
$$\bigg|\frac{x}{1+x} - \frac{c}{1+c}\bigg| = \bigg|\frac{x-c}{(1+x)(1+c)}\bigg| < \frac{\delta}{|(1+x)(1+c)|} = \frac{\delta}{|1+x|\cdot|1+c|}$$
Now, note that from the reverse triangle inequality, you can make that denominator smaller and thus yield a bigger expression, by :
$$1+x = x-c+1+c \Rightarrow |1+x| = |(x-c) - (-1-c)| \geq | |x-c| - |1+c|| =|δ-|1+c||$$
Thus, we get :
$$\bigg|\frac{x}{1+x} - \frac{c}{1+c}\bigg| = \bigg|\frac{x-c}{(1+x)(1+c)}\bigg| < \frac{\delta}{|(1+x)(1+c)|}\leq \frac{\delta}{|\delta-|1+c||\cdot|1+c|} \equiv \varepsilon$$
Summing it up, $\forall \varepsilon>0 \; \exists \delta >0 : |x-c|<\delta \implies \bigg|\frac{x}{1+x} - \frac{c}{1+c}\bigg| < \varepsilon$ and so the limit exists.
