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## #1 2012-11-07 18:49:37

princess snowwhite
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### finding limit

((1+cx)/(1-cx))^(1/x) tends to 4 as x tends to infinity.
find the limit of ((1+2cx)/(1-2cx))^(1/x) as x tends to infinity.

## #2 2012-11-07 19:41:28

bobbym

Online

### Re: finding limit

Hi princess snowwhite;

For the first part of the question:

((1+cx)/(1-cx))^(1/x) tends to 4 as x tends to infinity.

I can not find any c that will give 4 as x approaches infinity.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #3 2012-11-08 00:46:26

anonimnystefy
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### Re: finding limit

The first limit is 1 for every c.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #4 2012-11-08 05:21:06

bobbym

Online

### Re: finding limit

Hi;

That is what I am getting, so the question does not make any sense.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #5 2012-11-08 06:20:27

anonimnystefy
Real Member

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### Re: finding limit

Maybe it is a misprint or a typo.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #6 2012-11-08 06:27:03

bobbym

Online

### Re: finding limit

find the limit of ((1+2cx)/(1-2cx))^(1/x) as x tends to infinity

So the answer should be 1 because 2c=c.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #7 2012-11-08 07:25:20

anonimnystefy
Real Member

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### Re: finding limit

I don't think that is the misprint.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #8 2012-11-08 07:33:30

bobbym

Online

### Re: finding limit

But doesn't it follow?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #9 2012-11-08 07:36:30

anonimnystefy
Real Member

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### Re: finding limit

Not really. The question would be close to trivial. I think the question is more likely to be asking for the constant c for which a limit has the value 4 and then finding the same limit but with 2c instead of the c, whose value we now know.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #10 2012-11-08 07:40:18

bobbym

Online

### Re: finding limit

#### anonimnystefy wrote:

The first limit is 1 for every c.

There is no c that will give a limit of 4. You said so yourself. The limit is independent of c. That is why it is trivial.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #11 2012-11-08 07:45:48

anonimnystefy
Real Member

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### Re: finding limit

Yes. So I am thinking that it might be a typo of the function whose limit is being taken.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #12 2012-11-08 07:48:42

bobbym

Online

### Re: finding limit

Maybe the questioner just wants the OP to see that the answer is 1. To spot the inconsistency.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #13 2012-11-08 07:53:17

anonimnystefy
Real Member

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### Re: finding limit

This looks too much like a book problem. I do not think that is what it wants.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #14 2012-11-08 07:55:33

bobbym

Online

### Re: finding limit

Isn't the simplest typo that she wrote 4 when she meant 1?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #15 2012-11-08 08:00:10

anonimnystefy
Real Member

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### Re: finding limit

Being simplest doesn't makng true. And, yes, I am familiar with the Occam's razor.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #16 2012-11-08 08:01:57

bobbym

Online

### Re: finding limit

I use Schick's or Gillette's razor for a better shave.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #17 2012-11-13 01:26:50

mathteacher005
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### Re: finding limit

Hello! The answer is \lim_{x\to\infty }\left({{2\,{\it cx}+1}\over{1-2\,{\it cx}}}\right)^{{{1}\over{x}}} = 1.
You can check it here.
My friend show me few weeks ago and now my students always use it for checking the answers:) Good Luck!

## #18 2012-11-13 05:29:24

bob bundy
Moderator

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### Re: finding limit

hi mathteacher005

Welcome to the forum and thanks for the link.

I will convert your post so that the expression is in  Latex and the link easier to use.

numberempire.com/limitcalculator.php?fu … =two-sided

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei