How to prove this function is continuous on $(-1,1)$ How to prove $f(x)=\dfrac{x}{(1-x)^2}$ is continuous on $(-1,1)$, using either the $\epsilon$-$\delta $ definition or some other method. 
 A: Let $c \in ]-1, 1[$. If $x \in ]-1,1[$, then
$$
\bigg| \frac{x}{(1-x)^{2}} - \frac{c}{(1-c)^{2}} \bigg| = 
\frac{|x(1-c)^{2} - c(1-x)^{2}|}{(1-x)^{2}(1-c)^{2}} = \frac{|x-c||1-xc|}{(1-x)^{2}(1-c)^{2}};
$$
if in addition $|x-c| < 1$, then $|x|-|c| \leq |x-c| < 1$, implying $|x| < 1+|c|$, implying $|xc| < |c| + c^{2}$, implying $|1-xc| \leq |xc|+1 < |c|+c^{2}+1 =: M_{1,c}$, implying 
$$
\frac{|x-c||1-xc|}{(1-x)^{2}(1-c)^{2}} < \frac{|x-c|M_{1,c}}{(1-x)^{2}(1-c)^{2}};
$$
if in addition $|x-c| < |c-1|/2$, then 
$||x-1| - |c-1|| \leq |x-1-(c-1)| < |c-1|/2$, implying $|x-1| > |c-1|/2 =: M_{2,c}$, implying 
$$
\frac{|x-c|M_{1,c}}{(1-x)^{2}(1-c)^{2}} < \frac{|x-c|M_{1,c}}{M^{2}_{2,c}(1-c)^{2}};
$$
given any $\varepsilon > 0$, we have
$$
\frac{|x-c|M_{1,c}}{M^{2}_{2,c}(1-c)^{2}} < \varepsilon
$$
if in addition we have
$|x-c| < \varepsilon M^{2}_{2,c}(1-c)^{2}/M_{1,c}$.
All in all, for every $c \in ]-1,1[$ and every $\varepsilon > 0$ it holds that $x \in ]-1,1[$ and $|x-c| < \min \{ 1, |c-1|/2, \varepsilon M^{2}_{2,c}(1-c)^{2}/M_{1,c} \}$ imply
$|f(x) - f(c)| < \varepsilon$.
