Epsilon/Limits Help

**KarlosK** · February 10th 2010, 01:12 PM

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!

**Plato** · February 10th 2010, 01:45 PM

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} $

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