<s>[INST] 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. 
 [/INST] I like to my limits
go to zero,
so,
let $x = 1+y$.
Then
$\begin{array}\\
\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert
&=\lvert \frac{1}{(1+y)^3+(1+y)}-\frac{1}{2} \rvert\\
&=\lvert \frac{1}{y^3+3y^2+4y+2}-\frac{1}{2} \rvert\\
&=\lvert \frac{2-(y^3+3y^2+4y+2)}{2(y^3+3y^2+4y+2)}\rvert\\
&=\lvert \frac{-(y^3+3y^2+4y)}{2(y^3+3y^2+4y+2)}\rvert\\
&=\lvert \frac{-y(y^2+3y+4)}{2(y^3+3y^2+4y+2)}\rvert\\
\end{array}
$
As $y \to 0$,
$\frac{(y^2+3y+4)}{2(y^3+3y^2+4y+2)}
\to 1
$,
so
$\frac{-y(y^2+3y+4)}{2(y^3+3y^2+4y+2)}
\to 0$.
</s>