Prove directly: (x+1)^3 ----> 1 as x----> 0.
So 0<mod(x-0)< delta.
need help cheers
$\displaystyle \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3 \leqslant \left[ {\left| x \right| + 1} \right]^3 < 8$
$\displaystyle \varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}$
Not really. I was trying to show you how to think about it.
$\displaystyle \left| {\left( {x + 1} \right)^3 - 1} \right| = \left| x \right|\left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$
Now use the ideas in the other post to bound the factor $\displaystyle \left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$