1. ## Limits and Continuity

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

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

need help cheers

2. Originally Posted by ardam
Prove directly: (x+1)^3 ----> 1 as x----> 0.
$\displaystyle \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3 \leqslant \left[ {\left| x \right| + 1} \right]^3 < 8$

$\displaystyle \varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}$

3. Originally Posted by Plato
$\displaystyle \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3 \leqslant \left[ {\left| x \right| + 1} \right]^3 < 8$

$\displaystyle \varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}$
So you used the condition \delta < (and equal to) 1
To obtain [tex]\left| x \right| < 1\ which means x is in distance of one.
So by putting x = 1 into (x+1)^3 you got 8.
Is this correct?

4. Originally Posted by ardam
So you used the condition \delta < (and equal to) 1
To obtain [tex]\left| x \right| < 1\ which means x is in distance of one.
So by putting x = 1 into (x+1)^3 you got 8.
Is this correct?
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 $\displaystyle \left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$