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

In summary: So delta corresponds to epsilon when x-1<d.In summary, the conversation is about proving that x^3-5x+6-2=2. Prove is trying to find a delta that corresponds to epsilon =0.2, and then replace it with delta. Epsilon is 0.2.
  • #1
CrossFit415
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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
 
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  • #2
What do you mean by "epsilon = 2"? Are you supposed to find a suitable delta?
 
  • #3
Whoops I meant 0.2 Yea I'm trying to find delta.
 
  • #4
Since you somehow want to use the fact that |x-1| < [itex]\delta[/itex], you should be trying to divide this polynomial by x-1.
Then you'll have |x3-5x+4| = |x-1| |P(x) | < [itex]\delta[/itex] |P(x)|, where P(x) is some other polynomial of order 2, which you could also bound...

Try that. :)
 
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  • #5
I'm sorry but I still do not understand. Couldn't we just solve for epsilon then replace it for delta? Thanks
 
  • #6
CrossFit415 said:
Whoops I meant 0.2 Yea I'm trying to find delta.

CrossFit415 said:
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 [itex]\delta[/itex], so that for every x that holds |x-1| < [itex]\delta[/itex],
|f(x) - 2|| < 0.2.

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

1. What is the first step in solving this limit problem?

The first step in solving this limit problem is to set the given expression equal to the limit value, in this case 0.2. This will help us determine what value x needs to approach in order for the expression to be less than 0.2.

2. How do you simplify the given expression?

To simplify the given expression, we can factor out a common term, in this case (x-2). This will leave us with (x-2)(x^2+2x+3). We can then use the quadratic formula to solve for the remaining quadratic expression.

3. What is the significance of setting the expression equal to the limit value?

Setting the expression equal to the limit value helps us find the specific value of x that will make the expression approach the limit. This value is known as the limit point.

4. Can you use the squeeze theorem to solve this limit problem?

Yes, the squeeze theorem can be used to solve this limit problem. By finding two other expressions that are greater and less than the original expression, we can use the squeeze theorem to determine the limit point.

5. What is the final step in solving this limit problem?

The final step in solving this limit problem is to use the limit point that we have found to determine the range of values of x that will make the original expression less than 0.2. This will give us the solution to the limit problem.

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