
#1
Sep1011, 06:06 PM

P: 162

Prove; limit as x>1 (x^35x+6) = 2, epsilon=0.2
I got x^35x+62<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^35x+4 < 0.2 ? I'm on mobile can't use latex. Thanks 



#2
Sep1011, 06:13 PM

P: 202

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




#3
Sep1011, 06:19 PM

P: 162

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




#4
Sep1011, 06:30 PM

P: 202

Limit help, confused
Since you somehow want to use the fact that x1 < [itex]\delta[/itex], you should be trying to divide this polynomial by x1.
Then you'll have x^{3}5x+4 = x1 P(x)  < [itex]\delta[/itex] P(x), where P(x) is some other polynomial of order 2, which you could also bound... Try that. :) 



#5
Sep1011, 06:38 PM

P: 162

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




#6
Sep1011, 06:50 PM

P: 202

You wrote x^35x+62<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 x1 < [itex]\delta[/itex], f(x)  2 < 0.2. So, like I said, you have to somehow use the fact that x1 < [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. 



#7
Sep1011, 07:03 PM

P: 162

Well it asks me use the graph to find a number Delta.
If x1 < d then (x^35x+6)2<0.2 Then it tells me find Delta that corresponds to epsilon =0.2 Oh ok, Im starting to see it. 


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