Answer
For all $\epsilon>0$, exists $\delta>0$ so that
if $|x-0|
Work Step by Step
$f(x)=x^2+x^3$
We have:
Consider $|x|<1$
We have:
$|x^2+x^3-0|=|x^2(x+1)|$
$|f(x)-0|=|x^2(x+1)|0$, exists $\delta>0$ so that
if $|x-0|
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