Specific question on proving continuity of this function If $c\in\mathbb{R}$, I want to show that $f(x)=\dfrac{cx}{x^2+c^2}$ is continuous using $\varepsilon-\delta$. Unfortunately from an algebraic standpoint I seem to be missing the strategy. I let $p\in\mathbb{R}$ ($p\neq 0)$, and I want $\forall \varepsilon>0$ $\exists \delta>0$ such that $\left|x-p\right|<\delta \Rightarrow \left|f(x)-f(p)\right|<\varepsilon$.
$\left|f(x)-f(p)\right|=\left|\dfrac{cx}{x^2+c^2}-\dfrac{cp}{p^2+c^2}\right|=|c|\left|\dfrac{x}{x^2+c^2}-\dfrac{p}{p^2+c^2}\right|,$ but I do not see how to proceed.

Choose $\delta < |p|/2$ such that when $|x-p| < \delta$,
$$||x|-|p|| \leq |x-p|< |p|/2$$ 
and
$$|p|/2 <|x| < 3|p|/2$$
Note that
$$|f(x) - f(p)| < \frac{|c||p||x|}{|x^2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|x^2+c^2||p^2+c^2|}|x-p| \\ < \frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}|x-p|\\ = \left[\frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}\right] |x-p|$$
Choose  $\delta'$ such that the RHS is less than $\epsilon$ if $|x-p|< \delta'$: 
$$\delta' = \epsilon \left[\frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}\right]^{-1}$$
Therefore, if $|x-p| < \min(\delta, \delta')$ then $|f(x) - f(p)| < \epsilon$.
