Babynini
  • Babynini
Proving limits with delta..
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Babynini
  • Babynini
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Babynini
  • 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
  • Babynini
@jim_thompson5910 you free? :)

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zepdrix
  • 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
  • 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
  • zepdrix
Yes, we were trying to find a way to "deal with" the |x+4| :)
zepdrix
  • 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
  • Babynini
hrmm what you did looks pretty close to what we did in class with other examples.
jim_thompson5910
  • 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
  • Babynini
So then it comes out like: ...?
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zepdrix
  • 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
  • zepdrix
And then in part \(\large\rm ii)\) I would maybe make a connection here:\[\large\rm 5
Babynini
  • Babynini
ahh right right.
zepdrix
  • 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
  • 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
  • zepdrix
Assume \(\large\rm \delta<1\). Maybe that sounds a little nicer >.< I dunno
Babynini
  • Babynini
Dude I was just looking at that too. So redundant haha
zepdrix
  • 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
  • Babynini
Oh goshh, really? So you're pro then!
Babynini
  • Babynini
hahaha best description of these ever. But I like English, so it's ok xP
Babynini
  • Babynini
Is it all looking good now? :)
zepdrix
  • zepdrix
Nooo I'm not a pro :O noob. learning. Looking good? I think soooo.
zepdrix
  • 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
  • zepdrix
Maybe I'm being a little fussy :) But these words mean things
Babynini
  • 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
  • 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
  • Babynini
mhm! I am taking Calc 1 this quarter.
Babynini
  • Babynini
Okay! one last time!! xD
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Babynini
  • Babynini
oh goood, I hope it gets more fun haha
zepdrix
  • 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
  • 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
  • Babynini
I guess I could just write delta > 0 there though. Since I go back at the end and do that fancy "Therefore ..."
zepdrix
  • 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
  • Babynini
Would it be redundant/dumb to do both? o.o
zepdrix
  • 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
  • Babynini
Ah ok. Yep, it just helps me from getting lost haha
zepdrix
  • zepdrix
woah woah woah we're forgetting something very crucial
zepdrix
  • zepdrix
wow im glad i caught this before it was too late
zepdrix
  • zepdrix
\[\Huge \blacksquare\]
zepdrix
  • zepdrix
buhahaha
Babynini
  • Babynini
hahaha I will put it in right now!!
Babynini
  • Babynini
thanks!
zepdrix
  • zepdrix
yay team

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