# Homework Help: Limit help, confused

1. Sep 10, 2011

### CrossFit415

Prove; limit as x->1 (x^3-5x+6) = 2, epsilon=0.2

I got |x^3-5x+6-2|<0.2

Then I don't know where to go from there. Should I add 2 to 0.2 first or subtract 2 from 6 to get x^3-5x+4 < 0.2 ?

I'm on mobile can't use latex. Thanks

Last edited: Sep 10, 2011
2. Sep 10, 2011

### Tomer

What do you mean by "epsilon = 2"? Are you supposed to find a suitable delta?

3. Sep 10, 2011

### CrossFit415

Whoops I meant 0.2 Yea I'm trying to find delta.

4. Sep 10, 2011

### Tomer

Since you somehow want to use the fact that |x-1| < $\delta$, you should be trying to divide this polynomial by x-1.
Then you'll have |x3-5x+4| = |x-1| |P(x) | < $\delta$ |P(x)|, where P(x) is some other polynomial of order 2, which you could also bound...

Try that. :)

Last edited: Sep 10, 2011
5. Sep 10, 2011

### CrossFit415

I'm sorry but I still do not understand. Couldn't we just solve for epsilon then replace it for delta? Thanks

6. Sep 10, 2011

### Tomer

I don't exactly understand what you mean.

You wrote |x^3-5x+6-2|<0.2
Defining f(x)= x3 - 5x + 6
But it's not that you need to solve an inequality. What you need to do, formally, is to find a $\delta$, so that for every x that holds |x-1| < $\delta$,
|f(x) - 2|| < 0.2.

So, like I said, you have to somehow use the fact that |x-1| < $\delta$.
You start off by writing |x3 - 5x + 6 - 2|, and you start looking for ways to somehow express it with, apart from other things, $\delta$. You are looking to see how small this $\delta$ needs to be, so that |x3 - 5x + 6 - 2| is small enough.

You start off in the way I wrote. Otherwise - I don't understand your question.

7. Sep 10, 2011

### CrossFit415

Well it asks me use the graph to find a number Delta.

If |x-1| < d then |(x^3-5x+6)-2|<0.2

Then it tells me find Delta that corresponds to epsilon =0.2

Oh ok, Im starting to see it.