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
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 |x^{3}-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. :)
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)= x^{3} - 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 |x^{3} - 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 |x^{3} - 5x + 6 - 2| is small enough. You start off in the way I wrote. Otherwise - I don't understand your question.
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.