Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\frac{-(x-1)^2}{(x-1)^2+2(x-1)+2}| = \frac{(x-1)^2}{|(x-1)^2+2(x-1)+2|}|$
after which I get stuck at trying to implement triangle inequality to push in the modulo in the denominator to get $|x-1|$ terms to substitute .

You have\begin{align}\frac x{x^2+1}-\frac12&=\frac{-x^2+2x-1}{2x^2+2}\\&=\frac{-(x-1)^2}{2x^2+2}.\end{align}Now, you alwas have $2x^2+2\geqslant2>1$. So$$\left|\frac{-(x-1)^2}{2x^2+2}\right|<(x-1)^2\tag1$$and therefore, given $\varepsilon>0$, if you take $|x-1|<\sqrt{\varepsilon}$, it will follow from $(1)$ that$$\left|\frac x{x^2+1}-\frac12\right|<\varepsilon.$$
