How to prove this limit using epsilon delta method? i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.
$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$
My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon\implies\left|\frac{x^2-c^2}{1+c^2+x^2+x^2c^2}\right|<\epsilon\implies\left|\frac{(x-c)(x+c)}{1+c^2+x^2+x^2c^2}\right|<\epsilon$$
But i dont how to proceed after this?

$$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|=\left|\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right|\le \left|\frac{1+|x||c|}{(1+x^2)(1+c^2)} \right||x-c|$$
Suppose $c\not=0$, then we do some analysis first.
If we require $|x-c|<\frac{|c|}{2}$, then $|1+x^2|\ge 1+\frac{c^2}{4},|x|<\frac{3|c|}{2}$, that is $\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|\le M(c)|x-c|$, where $M(c)>0$ is a fixed constant. Hence it suffices to further choose $|x-c|<\frac{\epsilon}{M(c)}$. 
Combine above two parts together, let $\delta=\min(\frac{|c|}{2},\frac{\epsilon}{M(c)})$, and then $|x-c|<\delta\implies \left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon$
If $c=0$, you are proving $\lim_{x\to 0}\frac{x}{1+x^2}=0$, which is easier by squeezing theorem. 
