Showing $\frac{1}{x^3+x}$ is continuous at $x=1$ I've been attempting to use the definition of continuity and felt a little uncertain about my working on this question. 
We define $\epsilon>0$ and choose $\delta $ s.t ..... 
Let $\lvert x-1\rvert < 1 $: 
$\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert = \lvert \frac{2-x^3-x}{2x^3+2x}\rvert = \lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert$
Then we restrict $\delta < 1$ so now $ 0<x<2$ which tells us that 
$\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert<-\frac{2}{5}$
Hence $\lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert < -\frac{2\delta}{5}<\epsilon$. 
So we must choose $\delta > -\frac{2\epsilon}{5}$.
However I am used to finding something of the form $0<\delta<$min{$a,b$}
Does this mean the solution is  $-\frac{2\epsilon}{5}<\delta<1$ ? 
Apologies if this is really trivial/simple, I have just never seen a solution like that so wondered if perhaps I made an error somewhere. 

Note that we have
$$\left|\frac{1}{x(x^2+1)}-\frac12\right|=\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\,|x-1|$$
Now, we need to bound $x$ away from $0$.  So, let us first restrict $x$ so that $1/2<x<3/2$.  Then, we see that 
$$\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\le \frac{(3/2)^2+(3/2)+2}{((1/2)^2+1)}=\frac{23}{5}$$
Hence, we see that 
$$\left|\frac{1}{x(x^2+1)}-\frac12\right|\le \frac{23}5|x-1|<\epsilon$$
whenever $|x-1|<\delta=\min\left(1/2,\frac{5}{23}\epsilon\right)$.
