I understand (mostly) the solution to this problem however I'm having difficulty understanding how they've made the leap from $0<x<2$ to $0<|x||x+1|<6$ so if anyone could explain it to me I'd really appreciate it. Thanks

**Question**:

> 
> Give an $\epsilon$-$\delta$ proof of the continuity of the function $f(x)=x^3-x$ at $x=1$.
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> 
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**Answer**:

Note $f(1)=0.$ So given ϵ>0 want δ>0 s.t.
$|x-1|<δ => |x^3-x-0|<ϵ.$

I.e. 

$ |x(x^2-1)|<ϵ$

$|x||x+1||x-1|<ϵ$

Choose $\delta<1$ => 

$ 0<x<2 $. 
So, $0<|x||x+1|<6$.

So for $\delta< \min \{1,\frac\epsilon6\}$ we have $|f(x)-f(1)|<6(\epsilon/6)=\epsilon$.