# Given epsilon, find delta

1. Oct 23, 2008

### musicfairy

I don't know how yor format this, so:

x3 + x2 +x +1

The limit of that function = 0 as x approaches -1

What's the greatest value of delta when epsilon = 0.1?

This is what I tried to do:

|x3 + x2 + x + 1| < 0.1
-0.1 < x3 + x2 + x + 1 < 0.1
-1.1 < x3 + x2 + x < -0.9

In my calculator I plotted y = f(x), y = -1.1, and y = -0.9
The intersect are at x = -1.379 and x = -1.326

So I plugged them in this equation.

0<|x + 1|<delta

|-1.379 + 1| < 0.379
|-1.326 + 1| < 0.326

So I pick delta = 0.326 because it's smaller.

Now can someone please tell me everything that went wrong and offer any advice on how I could understand this?

2. Oct 23, 2008

### HallsofIvy

Staff Emeritus
What exactly was your f(x)? I do exactly what you say you did and I do NOT get -1.379 and -1.376. For one thing, x= -1 is NOT between those values- they can't possibly be right!

3. Oct 23, 2008

### Staff: Mentor

Your delta is way too big. f(-1.326) = -0.8992, and f(-0.674) = 0.474094. Here x is within .326 of -1 (-1 - .326 and -1 + .326), but the function values are not within .1 of zero.

When you plotted f(x) on your calculator, did you use f(x) = x^3 + x^2 + x + 1? You also show x^3 + x^2 + x in one of your inequalities, so I'm not sure whether you might have used that one.

You should factor x^3 + x^2 + x + 1 into (x + 1) times the other factor.

You want |x^3 + x^2 + x + 1| < epsilon
so |x + 1|* |something| < epsilon

You can tie that inequality to one involving |x + 1| if you replace |something| above with a value that's guaranteed to be larger, for all values of x on some reasonable interval.

As a hint, delta needs to be quite a bit smaller than epsilon, like some fraction of it.