<s>[INST] 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 .
 [/INST] As $ x $ goes to $ 1$, we can always assume that
$$|x-1|<2$$
and we look for $ \delta$ such that
$$|x-1|<2 \text{ and } |x-1|<\delta \implies$$
$$ |\frac{-(x-1)^2}{2x^2+2}|<\epsilon$$
but
$$2x^2+2\ge 2$$
and
$$|x-1|<2\implies $$
$$|\frac{(x-1)(x-1)}{2x^2+2}|\le |x-1|$$
So, we just need a $ \delta<2$ such that
$$|x-1|<\delta\implies |x-1|<\epsilon$$
from here, we see that we can take
$$\delta=\min(2,\epsilon)$$
</s>