Question

Use the limit definition to prove the limit
lim x->3(x^2 -5x +1)=-5

Answer 1

what definition are you using ?

Answer 2

are you using \(\epsilon, \delta\) business?

Answer 3

yess

Answer 4

k then the first part is a bunch of algebra

Answer 5

we want to show that given any \(\epsilon>0\) there is some \(\delta\) (which we write in terms of \(\epsilon\) so that if \(|x-3|<\dela\) we know that \(|x^2-5x+1+5|<\epsilon\)

Answer 6

typo there, i meant if \(|x-3|<\delta\) ...

Answer 7

can anyone help me plz too

Answer 8

okay so i get the first part but how do you choose

Answer 9

ughh...sorry still trying to get used to this

Answer 10

how do i choose delta

Answer 11

ok so are you good to this point? wanting to make \(|x^2-5x+1+5|<\epsilon\) ?

Answer 12

yeah, and then i get \[\left| x-2 \right|\left| x-3 \right|\]

Answer 13

oh then you are home free almost

Answer 14

you get control over the \(|x-3|\) part, i.e. you can make it as small as you like

Answer 15

now there is one little gimmick, what do to about the \(|x-2|\)

Answer 16

you have many choices, but since you get to say how close \(x\) is to \(3\), i.e. how small \(|x-3|\) is, you can say it is certainly less than \(1\)
if it is less than \(1\) that makes \(2<x<4\) in which case \(|x-2|\) is largest if \(x=4\) and \(|4-2|=2\) so at the very most \(|x-2||x-3|<2|x-3|\) if we assume \(|x-3|<1\)

Answer 17

and since if \(\delta<1\) we have \(|x-2||x-3|<2\delta\) all we have to do to insure it is less than \(\epsilon\) is to make \(|x-3|<\frac{\epsilon}{2}\)

Answer 18

does taking 1 work everytime, or just in this case?

Answer 19

since you have total control over \(|x-3|\) you could have picked \(|x-3|<5\) if you like, really makes no difference, i just picked \(1\) because it was simplest to compute with
and in any case we are considering small values of \(\epsilon\) not large ones

Answer 20

thank you soo much..this really helped..

Answer 21

you probably want to say if \(\delta<\text{min}(1,\frac{\epsilon}{2})\) then \(|x-2||x-3|<\epsilon\)

Answer 22

and if your teacher is a stickler for details you have to run the algebra backwards for the complete proof, but it is really exactly that, writing the steps in reverse order
yw

Answer 23

My book hardly has any answers, so even if I am doing something right I'm not really sure because I have no way to check..
My prof's alright..but I'm trying to work on my proofs still..

Answer 24

it is a whole new world, these proofs
but that is actually all of math, at least all of real math

Answer 25

Yeah this is the first time I'm doing proofs..don't like it at all!