# Epsilon/Limits Help

• Feb 10th 2010, 01:12 PM
KarlosK
Epsilon/Limits Help
I posted a question about these a few days ago and am still having trouble understanding this. If someone could show me how to do this problem with some explanation hopefully I can understand it better.

Find a value of alpha > 0 such that for all x, 0 < abs value (x- Xo) < alpha -> a<x<b.

a= -7/2 b= -1/2 Xo= -3

Our teacher barely does an examples in class, and I don't even understand what this problem is looking for. Thanks for help!
• Feb 10th 2010, 01:45 PM
Plato
Quote:

Originally Posted by KarlosK
Find a value of alpha > 0 such that for all x, 0 < abs value (x- Xo) < alpha -> a<x<b.
a= -7/2 b= -1/2 Xo= -3

$\left| {a - x_0 } \right| = \left| {\frac{{ - 7}}{2} + 3} \right| = \frac{1}{2}$
$\left| {b - x_0 } \right| = \left| {\frac{{ - 1}}{2} + 3} \right| = \frac{5}{2}$
The smaller of those two is $\frac{1}{2}$.
Therefore $\alpha =\frac{1}{2}$.

To see why consider:
$\left| {x - x_0 } \right| < \alpha \Rightarrow \left| {x + 3} \right| < \frac{1}
{2} \Rightarrow - \frac{1}
{2} < x + 3 < \frac{1}
{2} \Rightarrow \frac{{ - 7}}
{2} < x < - \frac{5}
{2} < - \frac{1}
{2}
$