Question

Proving limits with delta..

Answer 1

Gaah sorry it is sideways. I got stuck at the end there, not sure where to go.
The answer, I think, should be some
min{ ?}

Answer 2

@jim_thompson5910 you free? :)

Answer 3

Let delta be sufficiently small, say \(\large\rm \delta<1\).
Then, \(\large\rm |x-2|<1\)\[\large\rm -1< x-2<1\]Adding 6,\[\large\rm 5<\color{royalblue}{x+4<7}\]And I think we can do something withhhhh this. Hmm.
I'm just trying to figure out how we relate this to absolute value.\[\large\rm |x-2||x+4|<|x-2|\cdot 7<\epsilon\]

Answer 4

\[\large\rm |x-2|<\frac{\epsilon}{7}\]And then you have to choose your delta carefully,
depending on which is bigger, this epsilon/7 or the 1 restriction that we put on delta.\[\large\rm \delta=\min\left\{1,~~\frac{\epsilon}{7}\right\}\]

Answer 5

Yes, we were trying to find a way to "deal with" the |x+4|
:)

Answer 6

I was able to get this relationship: \(\large\rm 5<\color{royalblue}{x+4<7}\)
But not something like this: \(\large\rm a< |x+4|<b\)

I hope that isn't going to matter..
I'm trying to think about it :\ hmm

Answer 7

hrmm what you did looks pretty close to what we did in class with other examples.

Answer 8

@zepdrix I think as long as a > 0 and b > 0, then

\[\Large a < x+4 < b \implies |a| < |x+4| < |b|\]

Answer 9

So then it comes out like: ...?

Answer 10

For part \(\large\rm i)\) at the last line I think you want to write:\[\large\rm |x-2||x+4|<\epsilon\]I don't think we can relate this to delta just yet at that point. hmm.

Answer 11

And then in part \(\large\rm ii)\) I would maybe make a connection here:\[\large\rm 5<x+4<7\qquad\implies\qquad |x+4|<7\]

Answer 12

ahh right right.

Answer 13

And after that `"So" line` I would get rid of the `equals sign` on the next line.
It reads like, the epsilon from the previous line is equivalent to the next stuff.
Maybe just the word "Then," instead of "="

Answer 14

Oh also, maybe too much words at the start of \(\large\rm ii)\)
hehe

Let \(\large\rm \delta\) be \(\large\rm \delta<1\).
When I'm reading this in my head it reads like:
"Let delta be delta is less than 1."

Answer 15

Assume \(\large\rm \delta<1\).

Maybe that sounds a little nicer >.< I dunno

Answer 16

Dude I was just looking at that too. So redundant haha

Answer 17

These are tricky huh? :d
it's like combing Math and English class! lol

I'm taking a whole class on these types of problems :3 it's a pain sometimes.

Answer 18

Oh goshh, really? So you're pro then!

Answer 19

hahaha best description of these ever. But I like English, so it's ok xP

Answer 20

Is it all looking good now? :)

Answer 21

Nooo I'm not a pro :O noob. learning.
Looking good? I think soooo.

Answer 22

We don't want a therefore on the last line.
We gotta pick a delta first! :D

Use another word like ummm...

`Let` delta=min{stuff}.

Therefore,
0<|x-2|<delta implies |f(x)-L|<epsilon

or some nonsense like that, ya? :D

Answer 23

Maybe I'm being a little fussy :)
But these words mean things

Answer 24

Ah well I think you're doing excellently then! xP
oh gosh. What is all this fanciness haha. Nah I think it's good. Fussy is better than having it incomplete!

Answer 25

Are you near the beginning of a Calc 1 course? :o
That's when I remember first going over these types of problems.

Everything was super fun after this stuff.
Related rates, max/min problems, mean-value theorem,
lots of cool normal math stuff.

Answer 26

mhm! I am taking Calc 1 this quarter.

Answer 27

Okay! one last time!! xD

Answer 28

oh goood, I hope it gets more fun haha

Answer 29

I don't understand the first line of your proof:
\(\large\rm \delta=\qquad\qquad >0\)

Do you go back later and fill in that gap after you have found a delta or something?

Answer 30

oh, yes. That is how the prof does it. So now i'd go back and put in the
delta = min{1, e/7} in there. I just forgot to do that yet xD

Answer 31

I guess I could just write
delta > 0 there though. Since I go back at the end and do that fancy "Therefore ..."

Answer 32

Hmm that's a clever way to do that :)
I like that.

Yah either way should be ok I think, since you're stating your delta that you found at the end.

Answer 33

Would it be redundant/dumb to do both? o.o

Answer 34

You mean, would it be dumb to put \(\large\rm \delta=\left\{1,\frac{\epsilon}{7}\right\}>0\) at the top
since you already have it at the bottom? Nahhh seems fine.

Why are they numbered i) and ii)?
Is that just a way to stay organized?

Answer 35

Ah ok.
Yep, it just helps me from getting lost haha

Answer 36

woah woah woah we're forgetting something very crucial

Answer 37

wow im glad i caught this before it was too late

Answer 38

\[\Huge \blacksquare\]

Answer 39

buhahaha

Answer 40

hahaha I will put it in right now!!

Answer 41

thanks!

Answer 42

yay team