Proving limits with delta..

- Babynini

Proving limits with delta..

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Babynini

##### 1 Attachment

- Babynini

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{ ?}

- Babynini

@jim_thompson5910 you free? :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- zepdrix

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.\[\large\rm |x-2||x+4|<|x-2|\cdot 7<\epsilon\]

- zepdrix

\[\large\rm |x-2|<\frac{\epsilon}{7}\]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.\[\large\rm \delta=\min\left\{1,~~\frac{\epsilon}{7}\right\}\]

- zepdrix

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

- zepdrix

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|

- Babynini

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

- jim_thompson5910

@zepdrix I think as long as a > 0 and b > 0, then
\[\Large a < x+4 < b \implies |a| < |x+4| < |b|\]

- Babynini

So then it comes out like: ...?

##### 1 Attachment

- zepdrix

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.

- zepdrix

And then in part \(\large\rm ii)\) I would maybe make a connection here:\[\large\rm 5

- Babynini

ahh right right.

- zepdrix

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 "="

- zepdrix

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."

- zepdrix

Assume \(\large\rm \delta<1\).
Maybe that sounds a little nicer >.< I dunno

- Babynini

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

- zepdrix

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.

- Babynini

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

- Babynini

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

- Babynini

Is it all looking good now? :)

- zepdrix

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

- zepdrix

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|

- zepdrix

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

- Babynini

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!

- zepdrix

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.

- Babynini

mhm! I am taking Calc 1 this quarter.

- Babynini

Okay! one last time!! xD

##### 1 Attachment

- Babynini

oh goood, I hope it gets more fun haha

- zepdrix

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?

- Babynini

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

- Babynini

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

- zepdrix

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.

- Babynini

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

- zepdrix

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?

- Babynini

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

- zepdrix

woah woah woah we're forgetting something very crucial

- zepdrix

wow im glad i caught this before it was too late

- zepdrix

\[\Huge \blacksquare\]

- zepdrix

buhahaha

- Babynini

hahaha I will put it in right now!!

- Babynini

thanks!

- zepdrix

yay team

Looking for something else?

Not the answer you are looking for? Search for more explanations.