Prove directly: (x+1)^3 ----> 1 as x----> 0.

So 0<mod(x-0)< delta.

need help cheers

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- Oct 12th 2009, 01:45 PMardamLimits and Continuity
Prove directly: (x+1)^3 ----> 1 as x----> 0.

So 0<mod(x-0)< delta.

need help cheers - Oct 12th 2009, 01:54 PMPlato
$\displaystyle \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3 \leqslant \left[ {\left| x \right| + 1} \right]^3 < 8$

$\displaystyle \varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}$ - Oct 12th 2009, 02:05 PMardam
- Oct 12th 2009, 02:38 PMPlato
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 $\displaystyle \left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$