## Formal definition of the limit

1. The problem statement, all variables and given/known data
prove the statement using epsilon delta definition of the limit

lim x->2 (x^3)=8
2. Relevant equations
lim x->a f(x)=L is true when
|f(x)-L|<e
whenever
0< |x-a|<d

3. The attempt at a solution

|x^3 - 8| < e
|(x-2)| |(x^2+2x+4)|<e
x^2+2x+4 has no real roots there for only has one sign f(0)=4 so it is always positive
so |x^2+2x+4|=x^2+2x+4
|(x-2)||(x^2+2x+4)| <C(x-2)<e
where
|(x^2+2x+4)|<C
(x-2) < (e/C) =d

I usually solve most problems like this but I can't find the C
I know that we usually say that d is a small distance so |x-a| <1
|x-2|<1
1<x<3
but then what exactly?
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 Recognitions: Homework Help don't you want to find C such that $|(x^2+2x+4)| \leq C$ for all x?
 yes I am asking how to start that. also Does it have to be for all x? I mean aslong as I can find a C that works for x that are near two it is enough

## Formal definition of the limit

how can I relate or change |x-2|<1 to something like x^2+2x+4<C?
if I square it
x^2-4x+4<1
x^2+2x+4<1+6x
I will get a variable not a constant
if I cube it
(x-2)(x-2)^2<1
x^2-4x+4<1/(x-2)
x^2+2x+4<6x+ 1/(x-2)
x^2+2x+4< (6x^2-12x+1)/(x-2)
x^3-8<6x^2-12x+1
this goes in rounds too
I don't know If I got this right but if d is less than 1 then the maximum value of x that can be inputted into x^2+2x+4 is 3
and cant I get C to be (6x^2-12x+1)/(x-2) of 3 = 19
so can't I just say that as long as delta <1 and e/19 the limit is true?
Also what I don't understand about these questions are what delta is necessary for it to be true?I mean I just speculated the solution but what does it mean that delta has to be less than e/19?
 Is it because I didn't show enough work that I don't get any replies?
 Recognitions: Homework Help so we need to show that for any given $\epsilon> 0$ we can choose a $\delta > 0$ such that for any x satisfying $|x-2| < \delta$ then $$|x^3 - 8 | < \epsilon$$ so expanding as you have $$|x - 2 |.|x^2+2x+4| < \epsilon$$ now it should be clear (apologies in the other post I put a less than) $$x^2+2x+4 \geq 3$$ thus we have $$3|x - 2 | \leq |x - 2 |.|x^2+2x+4| < \epsilon$$ hopefully you can take it from here
 Recognitions: Homework Help another way to do it would be to work in terms of a variable say d, such that $-\delta < d < \delta$, then you can write the limit as $$|(2+d)^3 - 8| < \epsilon$$ expanding that should allow you to solve for d, making the requirerd relation between epsilon & delta clearer
 3|x-2|
 Recognitions: Homework Help that'd do it the other way was $$|(2+d)^3 - 8| < \epsilon$$ $$|(2+d)(4+4d +d^2) - 8| < \epsilon$$ $$|(8+8d+2d^2+4d+4d^2+d^3) - 8| < \epsilon$$ $$|(12d+6d^2+d^3)| < \epsilon$$ $$|d| |(12+6d+d^2)| < \epsilon$$ $$|d| |(12+6d+d^2)| < \epsilon$$ similarly $$|(12+6d+d^2)| \geq 3$$

Recognitions:
Homework Help
 Quote by lanedance $$|d| |(12+6d+d^2)| < \epsilon$$
We prove that the limit is 2 if - for any epsilon - a value for d is found so that |x^3-8|<epsilon if |x-2|<d. Unfortunately, we have a complicated expression for d. But remember, both epsilon and d are small numbers, much smaller than 1. So 12+6d+d^2 is certainly overestimated by replacing d with 1. So it is quite sure that d= epsilon/19 will do.

ehild
 Recognitions: Homework Help not that it changes the idea, but why go to e/19, when you can show it with the looser constraint e/3?
 Recognitions: Homework Help To get d you need the stricter constraint. Just try. Let be e=0.1. What should be d so as |(2+d)^3-8|
 Recognitions: Homework Help yeah apologies, miscalulation on my part
 I did figure what epsilon is but I have problems writing it as a formal proof any hints about that?
 Recognitions: Homework Help You have to find out d can be if you know epsilon. As the definition of the limit is: "lim x->a f(x)=L is true when |f(x)-L|
 Given e>0 we choose d =min{1,e/19} such that to show that 0<|x-2|-1,.'. f(1)2 x^3=8 --------------------------------------------- What are the unnecessary details? what should be added or deleted? or should I do a completely different format?

Recognitions:
Homework Help
 Quote by madah12 Given e>0 we choose d =min{1,e/19} such that to show that 0<|x-2|
It is the opposite: Given arbitrary small e>0 we choose d such that whenever 0<|x-2|<d, |x^3-8|<e. It is all right otherwise.

ehild