# Homework Help: Prove lim x/(x+1) = 1/2

1. Feb 26, 2010

### kingwinner

1. The problem statement, all variables and given/known data
Prove
lim x/(x+1) = 1/2
x->1

2. Relevant equations

3. The attempt at a solution
|x/(x+1) - 1/2|=|x-1|/(2|x+1|)
Assume x>0 (can I say this??), then |x-1|/(2|x+1|)<|x-1|/2
Take delta=min{1,epsilon/2}.
Then if 0<|x-1|<delta, then |x/(x+1) - 1/2|<epsilon

In the middle of my proof, I assumed that x>0, is this OK?
Does my choice of delta (delta=min{1,epsilon/2}) work?

Could someone kindly confirm this? (or point out any mistakes)
I haven't done those for awhile, so I'm not sure if I'm doing it correctly.
Thanks!

2. Feb 26, 2010

### tiny-tim

Hi kingwinner!

(have a delta: δ and an epsilon: ε )

I don't like your x > 0 … it's messy, and you really ought to do x < 0 also.

You're trying to get a lower limit on |x+1|, and you've chosen 2 … which doesn't really work for x < 0, does it?

try something less than 2, eg 1.5

3. Feb 26, 2010

### kingwinner

hmm...so my "proof" above is wrong? :(

But I thought the definition of limit says as x gets close to 1 (0<|x-1|<delta), then the function gets arbitrarily close to the limit 1/2. We don't really have to care at all about the case x<0, right? If I take delta=min{1,epsilon/2}, then delta is always less than or equal to 1, so we never have to look at the case x<0, isn't it?

So if I say I assume x>0 in the first place of the proof. Is this ok, can somebody confirm or correct me, please?

Last edited: Feb 26, 2010
4. Feb 26, 2010

### arildno

tiny-tim!

I totally disagree with you!

Being very close to x=1 implies x>0, so there is no need to consider x<0.

kingwinner:

Under your assumption x>0 (perfectly okay!), you have derived the bound <|x-1|/2

Thus, by chosing delta = 2e, whenever |x-1|<delta, your expression will be less than e.

5. Feb 26, 2010

### kingwinner

x>0 => 1- delta > 0 => delta <1 so I think here is the requirement in addition to the bound I obtained later on.

So I think if I take delta=min{1, 2e}, then both conditions hold, and if 0<|x-1|<delta, then |x/(x+1) - 1/2|<epsilon ? I don't think it's enough to just take delta=2e...we need delta to be no greater than 1 as well, right?

6. Feb 26, 2010

### tiny-tim

Hi arildno!

Yes, you're right, I was confused about what x was (also I missed the factor of 2 in 1/2|x+1|).

Thanks for the correction.

7. Feb 26, 2010

### kingwinner

I actually originally meant to take delta=min{1, 2e}, but I made a silly computational mistake and got e/2 instead of 2e...

But if I really take delta=min{1, e/2}, it's going to work as well because if a certain delta works in the definition, than any smaller delta (>0) will certainly work, right?

8. Feb 26, 2010

Absolutely!