Solving a Limit Problem: x^3-5x+6-2 < 0.2

[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

 

[Tomer]

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

 

[CrossFit415]

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

 

[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. :)

 

[CrossFit415]

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

 

[Tomer]

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

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

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.

 

[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, I am starting to see it.