# Limits and Continuity

• Oct 12th 2009, 01:45 PM
ardam
Limits 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 PM
Plato
Quote:

Originally Posted by ardam
Prove directly: (x+1)^3 ----> 1 as x----> 0.

$\left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3 \leqslant \left[ {\left| x \right| + 1} \right]^3 < 8$

$\varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}$
• Oct 12th 2009, 02:05 PM
ardam
Quote:

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

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

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?
• Oct 12th 2009, 02:38 PM
Plato
Quote:

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.
$\left| {\left( {x + 1} \right)^3 - 1} \right| = \left| x \right|\left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$

Now use the ideas in the other post to bound the factor $\left| {\left( {x + 1} \right)^2 + \left( {x + 1} \right) + 1} \right|$