Epsilon Delta Proof

• Dec 8th 2007, 06:39 PM
akhayoon
Epsilon Delta Proof
$\displaystyle \lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}$

apparently $\displaystyle Delta=\frac {3Epsilon}{5}$

you see I know how they got the 5

but for the 3 all they did was say $\displaystyle |x^2+x+1|>1$

and for some reason that lead them to 3

MAT196F Midterm Test Solutions

it's kind of hard to read but it's the first question
• Dec 8th 2007, 06:55 PM
Jhevon
Quote:

Originally Posted by akhayoon
$\displaystyle \lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}$

apparently $\displaystyle Delta=\frac {3Epsilon}{5}$

you see I know how they got the 5

but for the 3 all they did was say $\displaystyle |x^2+x+1|>1$

and for some reason that lead them to 3

MAT196F Midterm Test Solutions

it's kind of hard to read but it's the first question

no. for the 3, they combined the fractions. that is, they subtracted 2/3 from the expression
• Dec 8th 2007, 07:02 PM
akhayoon
I know they did that

but they said that if $\displaystyle |x-1| < 1, then 0 < x < 2, so, |2x+1| < 5$

so this give you permission to turn the |2x-1| into a five but to get rid of the

$\displaystyle x^2+x+1$ all they said is that it's bigger than one so there

so how does that give them permission to make the that polynomial equal 1
??
• Dec 8th 2007, 07:17 PM
Jhevon
Quote:

Originally Posted by akhayoon
I know they did that

but they said that if $\displaystyle |x-1| < 1, then 0 < x < 2, so, |2x+1| < 5$

so this give you permission to turn the |2x-1| into a five but to get rid of the

$\displaystyle x^2+x+1$ all they said is that it's bigger than one so there

so how does that give them permission to make the that polynomial equal 1
??

i'm sorry, i don't see what you are talking about. where did they equate the polynomial to 1? they just said |x^2 + x + 1| > 1. which is okay.

This is my 57:):)th post!!!!
• Dec 8th 2007, 07:28 PM
akhayoon
I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these:)

$\displaystyle \frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E$

|2x-1|<5 so that means that I could use 5 instead of this expression
I know they got this from $\displaystyle 0<x<2$ sweet...

and then they go $\displaystyle |x^2+x+1| > 1$
but where did this come from did they use $\displaystyle |x-1|<1$ again somehow?? :eek:
• Dec 8th 2007, 07:48 PM
Jhevon
Quote:

Originally Posted by akhayoon
I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these:)

$\displaystyle \frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E$

|2x-1|<5 so that means that I could use 5 instead of this expression
I know they got this from $\displaystyle 0<x<2$ sweet...

and then they go $\displaystyle |x^2+x+1| > 1$
but where did this come from did they use $\displaystyle |x-1|<1$ again somehow?? :eek:

yes, but not directly. they got that from 0 < x < 2. note that when x = 0, the expression |x^2 + x + 1| = 1. but x is greater than zero, so we must have |x^2 + x + 1| > 1, since we increased the expression by a positive amount

are you still confused?
• Dec 8th 2007, 07:56 PM
akhayoon
ok thanks, that explains it, so why couldn't I just make

|2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?

why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?:(
• Dec 8th 2007, 08:13 PM
Jhevon
Quote:

Originally Posted by akhayoon
ok thanks, that explains it, so why couldn't I just make

|2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?

why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?:(

remember, we stated out with |x - 1|< 1, which is why things played out that way. besides, if x = 0, |2x + 1| is not bigger than 1. we need to make x a range of values here, as we wanted inequalities. the actual value doesn't matter that much. we could have made |x - 1| < 1/2 or whatever. but 1 looks nice