Question

So I know that the lim (f(x) + g(x)) = lim f(x) + lim g(x).

I was wondering if the converse is true. If so, how do I prove it. Thanks!

Answer 1

Yes, there is a way to prove that.
Can I show you by an example?

Answer 2

Sure. Thanks.

Answer 3

now let's do the left side

Answer 4

Thanks. That makes sense. Do you know if there is a formal proof for the converse? I know the proof for the one above, I'm just stumped on how to set up the proof for the converse.

Answer 5

Kind of. I was wondering if there was a formal proof for this?

Answer 6

Like is there a delta/episilon proof for the converse or some other type of proof for the converse. I know the delta/episilon proof for the original, just not sure how to approach the converse.

Answer 7

I'll keep looking out for more formal proofs of this limit. I'll message you if I find more.

Answer 8

I've looked all over the internet and can't find anything. Thanks for your help!

Answer 9

Daw, but there are some peeps on Openstudy that is very resourceful. You can try asking them too.

Answer 10

or him*

Answer 11

Thanks, I appreciate your help.

Answer 12

He might be busy, sleeping I think. I can try asking him later.
He's the guy that always have a lot of good resources.

Answer 13

You're welcomed :) see you around soon.

Answer 14

thank you @xapproachesinfinity

Answer 15

@xapproachesinfinity the top is me proving it wrong. it's actually involved episilon and delta to prove the limit, which is beyond my level of understanding.

from what I know this the college level of calc, and he's a math major taking his college final exam in the next friday.

Answer 16

Btw, your profile pic looks really yummy

Answer 17

hmm, I'm not sure how to go around the delta epsilon proof.
the delta epsilon states that if |x-a|<delta then |f(x)-L|<epsilon
may be you need to set up this definition to two functions to prove the converse.

Answer 18

haha that is called tagine lol

Answer 19

what are your thoughts? eru?

Answer 20

Thanks xapproach :) Idk him well but I thought this is important because it would affect his math major career so.

Thank you so much for coming in. I tried asking alot of people >.<

Answer 21

I guess i haven't seen a poof to this before! I'm kinda rusty with proofs now lol.
can we set up something to work with like
let L1=lim f as x goes to c
and L2=lim g as x goes to c

Answer 22

hm, do you think this is it? I don't know how to read this type of proof so I'm not sure if this is converse or not.
http://math.stackexchange.com/questions/408135/prove-limsup-limits-n-to-infty-a-nb-n-le-limsup-limits-n-to-infty?rq=1

Answer 23

especially when they are using a sub n , b sub n
not f(x) g(x)

Answer 24

those are supremum proof! I haven't studied that as far as i remember.
that's dealing with ap, gp

Answer 25

may be the property |a+b|<|a|+|b| will be of a help here! not sure

Answer 26

oh I have ZERO clue what episilon mean xD left alone ap , gp. You're already really good!
This type of math is just over our knowledge >.<

Answer 27

Arithmetic progression and geometric progression. you're not familiar with those

Answer 28

@dan815 help here please!

Answer 29

oh so that's what it means. Yes I'm familiar, but the calc I'm dealing with doesn't include using episilon so.

so we just give up? : o

Answer 30

hehe, there sure will be someone to help! I familiar with thee epsilon definition
but I forget what i knew lol. that's why i'm starting over

Answer 31

@SithsAndGiggles please help with this!

Answer 32

oh, good luck! I hope someone finds the answer to this v.v tried for 4 hours already xD

Answer 33

thanks again xapproaches. You're so nice :))

Answer 34

I haven't done anything lol, but anytime!

Answer 35

I want to see how to get the answer to this too now haha

Answer 36

|x-c|<deltha1==>|f(x)-L1|<epsilon1
|x-c|<deltha2==>|f(x)-L2|<epsilon2
can we use any property here?

Answer 37

We assume that L1 and L2 exist for f and g respectively

Answer 38

the second line I meant g(x)

Answer 39

If we try, we'll get it eventually : D

Answer 40

so far I'm understanding what you're proving :D

Answer 41

This from some of limits to limit of sum so it the converse? that's what we are looking for

Answer 42

Oh good! I need a hand for this lol

Answer 43

can we add the two statements? not sure if it is correct though

Answer 44

like 2|x-c|<deltha1 +deltha 2 ==> |f(x)-L1|+|g(x)-L2|< epsilon1 + epislon2
this weird not sure if it is logical

Answer 45

may be 2|x-c|<deltha1 + deltha2 is good to verify but the implication we need someone to check this haha

Answer 46

oh god xD sometimes I really hate myself for not knowing things. Let me tag one more person to check--- yes, we should ask!

Answer 47

maybe nincompoop?

Answer 48

Tagg him/her!

Answer 49

@nincompoop please help check this proof? Thank you Nin

Answer 50

Really hope he comes.

Answer 51

|f(x)-L1|+|g(x)-L2|> |f(x)+g(x)-(L1+L2)|

Answer 52

this seem the flaw little bit haha

Answer 53

Now we got |f(x)+g(x)-(L1+L2)|<epsilon1 + epsilon2

Answer 54

we are half way now?

Answer 55

@ganeshie8 can you check this please! I'm sure I'm doin' this right

Answer 56

not*

Answer 57

I think we are xD but not sure if we should go further

Answer 58

until someone comes and checks

Answer 59

@tkhunny please help check this proof, we are so unsure of its accuracy

Answer 60

if |x-c|<(deltha1+deltha2)/2 holds as we said above then it implies
|(f(x)+g(x))-(L1+L2)|<epsilon1+epsilon2
can we deduce that the limit of (f(x)+g(x)) exist and it is L1+L2?
or may be we need further steps!
based on the epsilon definition this should be good, assuming the above step are true!

Answer 61

Yes! All we need is someone clarifying this.

Answer 62

Thanks for your help!

Answer 63

Just curious, ehub, how did you do it?

Answer 64

and is our way of doing this proof correct?

Answer 65

@ehubbard0307, you are welcome! not sure this is the answer? please wait for ppl who have more resources on this

Answer 66

looks fine to me?

Answer 67

@ash2326

Answer 68

@ehubbard0307 have you got your answer to this question yet?
If yes, I hope you don't mind sharing it with us, or perhaps check our proof?

I'm not sure of what your message means >.<

Answer 69

@aum, see this please!

Answer 70

I'm coming back!

Answer 71

Ok!

Answer 72

the sleep will not come if this is left unsolved haha

Answer 73

Consider f(x) = x and g(x) = -x

Answer 74

Assuming\[\begin{cases}\lim_{x\to c}f(x)=L\\\lim_{x\to c}g(x)=M\end{cases}\]
you want to show that
\[\lim_{x\to c}\bigg(f(x)+g(x)\bigg)=L+M\]
For the proof, you want to find \(\delta\) such that \(0<|x-c|<\delta\) implies \(\left|\bigg(f(x)+g(x)\bigg)-\bigg(L+M\bigg)\right|<\epsilon\).

To do that, you work backwards with the previous inequality and rewrite as needed:
\[\begin{align*}\left|\bigg(f(x)+g(x)\bigg)-\bigg(L+M\bigg)\right|&=\left|\bigg(f(x)-L\bigg)+\bigg(g(x)-M\bigg)\right|&(1)\\
&\le |f(x)-L|+|g(x)-M|&(2)\\
&<\frac{\epsilon}{2}+\frac{\epsilon}{2}&(3)\\
&<\epsilon
\end{align*}\]
(1) rewrite the expression
(2) triangle inequality
(3) we know the individual limits exist if we choose to take the least of the two \(\delta\)s for the known limits. In other words, we're given \(0<|x-c|<\delta_1\) for \(f\) and \(0<|x-c|<\delta_2\) for \(g\) - we pick \(\delta=\min\{\delta_1,\delta_2\}\).

Answer 75

Thank you!

Answer 76

@SithsAndGiggles, thanks for the proof
can we say a ==> b
and c==>d
then a+c ==>b+d

Answer 77

@ganeshie8

Answer 78

Have to object to Siths very clean demonstration. The very first statement is that the limits exist. Without the assumption that the limits exist, it is an entirely different story.

Please consider the two functions I suggested and don't make the assumption of existing limits.

Answer 79

Well he assumed the limits exist in the first line too? didn't he
or may be im understanding it in a wrong way.Thanks for checking this

Answer 80

That is what I said, "The very first statement is that the limits exist."

Well, it's not perfectly obvious that this is the first statement, but it is clear later that it is the intent. A perfectly clear statement would have been:

\(\lim\limits_{x\rightarrow c}\;f(x) = L < \infty\)

This says rather deliberately that the limit exists and is finite.

Answer 81

@sithsandgiggles Are you sure that this would work for proving the converse of lim (f(x) + g(x)) = f(x) + g(x). It looks like the straightforward proof that I know. I know what a converse is, I'm just clueless on setting up a proof for it. I also have to prove the converse of lim(f(x)*g(x)) = limf(x)* limg(x).

Thanks for your help!

Answer 82

The statement does not have a converse because it is not an implication. It is an equality. And an equality works both ways.

Answer 83

It also doesn't have a converse because it isn't true at all as it is stated. Counter examples are trivially observed.

Answer 84

Everything you need is available here: http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx

Notice the language used in the statements proved in the link.

Answer 85

Thanks so much for your help guys! I get what you all are saying, I'm just confused more about the question itself.

Answer 86

Really, it is this:

1) If you have a counter example, you are done.
2) If you construct a valid proof, you are done.
3) Additional assumptions may be added in order to gain greater insight and to prove lesser statements.
4) Fewer assumptions are the sort of thing that advanced studies are about - The Generalization! It is generally agreed that a generalization should have at least two examples. :-)