Question

Using Epsilon - Delta definition, prove that the lim (x->1) X^2 + 3 = 4

Answer 1

So a better copy of the equation is
\[\lim_{x \rightarrow 1} x^{2} + 3 = 4\]

Answer 2

I wonder if this works the same way as our epsilon proofs. We do epsilon proofs, but we dont do them the traditional epsilon-delta way that was shown in calc 1, we kind of put a twist on them in our analysis class.

Answer 3

I know given Epsilon > 0, and Delta > 0
(x-1) => (\[|x^{2} + 3 - 4| < Epsilon\]

Answer 4

I'm not to sure @Concentrationalizing I'll show you what I go so far

Answer 5

\[|x ^{2}+3-4| => |x ^{2}-1| => |(x+1)(x-1)|\]

Answer 6

This is where I get my problem, on all the examples we can pull out the (x-1) but since it's being multiplied by (x+1) i'm not sure what to do

Answer 7

|(x+1)(x-1)| = |x+1| * |x-1|

Answer 8

thats an easy step up - the tricky part is cooking up a delta

Answer 9

agreed since |x-1| < delta, what do I do with the x+1

Answer 10

Yeah, ive seen something like this in a calc 1 textbook where they did something funky. But with my analysis class, we never went with the usual epsilon-delta idea, so Im not sure if what our class would do would be appropriate for yours. But in the example I saw, what was done was if you take an interval about your limit of 4, so let's say (3,5). In that interval, the extra part of your absolute value, the x+1, is less than 6, so for that reasoning, the inequality is then rewritten as:
\[6|x-1| < \epsilon \implies |x-1| <\epsilon /6\]

Answer 11

@Concentrationalizing I remember seeing something like that in my notes, let me check that out.

Answer 12

\(\large |x+1| |x-1| \lt \epsilon \implies |x-1| \lt\dfrac{\epsilon}{|x+1|}\)

Answer 13

Found an example in my notes, It states to let Delta = 1; => \[|x-1|< 1 \rightarrow -1<x-1<1 \rightarrow -1+2<x-1+2<1+2 \rightarrow 1<x+1<3\]

Answer 14

Therefore |x+1| < 3

Answer 15

They actually teach this in school?

Answer 16

haha Yup, It's in my real analysis class

Answer 17

This is the first time I've seen someone ask this, most profs skip over this.
Very cool though, I did some reading on this on my own haha.

Answer 18

It could be that my prof is pretty old school.

Answer 19

Haha, well that could be a benefit!

Answer 20

Nah, this is typical in analysis as far as Ive seen. Maybe skipped over or only touched on briefly in calc 1, but drilled into our heads in analysis xD

Answer 21

You're probably right

Answer 22

Hey @Concentrationalizing where you studying at?

Answer 23

UNLV. Born and raised here, so this is just the local university. If I had the means Id be elsewhere, lol.

Answer 24

Man I can only imagine the temptation not to study in Sin City

Answer 25

I think this should be taught in calc 1 as this is the precise definition of a limit

Answer 26

Anyways did you figure out what you needed?

Answer 27

Yes I did batman,

Answer 28

Right on :)

Answer 29

\(\large \delta\lt 1 \implies |x-1| \lt 1 \implies |x+1| \lt 3\)

nice :)

Answer 30

Thanks

Answer 31

That seems like a nice technique. Just pick a delta and use it to define the extra piece of the inequality for you.

Answer 32

however that only works when \(\large \epsilon \lt 3\), you need to tweak it further for \(\large \epsilon \gt 3\)

Answer 33

I'm not done, that just fixed my issue with the |x+1| I still have a little more work to do

Answer 34

if you want your function to be with in 100000, your delta wont work..

Answer 35

or it may work, need to think.. but yeah the proof is incomplete