Another Delta-Epsilon Question

[zooxanthellae]

Homework Statement

Prove that lim f(x) x --> 0 = lim f(x-a) x --> a.

Homework Equations

0 < |x-a| < d
0 < |f(x)-a| < E

The Attempt at a Solution

Spivak's logic, again, is giving me trouble. However, I realize that his method probably has a better chance of being right than mine. His method follows:

Suppose that lim f(x) x --> a = l, and let g(x) = f(x-a). Then for all E > 0, there is a d > 0 such that, for all x, if 0 < |x-a| < d, then |f(x)-l| < E. Now, if 0 < |y| < d, then 0 < |(y+a) - a| < d, so |f(y+a)-l| < E. But this last inequality can be written |g(y) - l| < E. So lim g(y) y --> 0 = l.

The bold portion is where I get lost. How can he rewrite |f(y+a)-l| < E as |g(y)-l| < E? Doesn't this suggest that f(y+a) = g(y)? However, wouldn't that conflict with the original definition of g(x) = f(x-a)? Wouldn't this definition mean g(y) = f(y-a), not f(y+a)? Where's my error?

 

[xaos]

if its true for all x, its true for all y such that y=x-a for some fixed a, then we relabel the hypothesis on x with y+a. (hopefully if I'm interpreting your problem correctly)

 

[zooxanthellae]

if its true for all x, its true for all y such that y=x-a for some fixed a, then we relabel the hypothesis on x with y+a. (hopefully if I'm interpreting your problem correctly)

I think you are, but I don't understand how the bold part is possible.

 

[snipez90]

Hmmm it does look a bit fishy, even though it really shouldn't. Anyways I think it's a lot easier to start with lim f(x) as x -> 0, and then simply replace x with the obvious substitution. I mean that's really the intuitive idea. If x is going to 0, might as well replace it with x-a, and let x go to a instead. This kind of thing comes up a few times, e.g. characterizations of continuity, equivalent definitions of derivative, and so on.

 

[xaos]

yes, i don't think this is Spivak's exact reasoning. you're assuming incorrectly either 'x' is near 'a' (it should be 'zero') or lim f(x-a) x-->a followed by g(x)=f(x-a), which is a little backwards and doesn't follow your reasoning.

 

[zooxanthellae]

Hmmm it does look a bit fishy, even though it really shouldn't. Anyways I think it's a lot easier to start with lim f(x) as x -> 0, and then simply replace x with the obvious substitution. I mean that's really the intuitive idea. If x is going to 0, might as well replace it with x-a, and let x go to a instead. This kind of thing comes up a few times, e.g. characterizations of continuity, equivalent definitions of derivative, and so on.

The thing is, Spivak specifically asks for a rigorous proof, which I'm assuming must mean deltas and epsilons.

yes, i don't think this is Spivak's exact reasoning. you're assuming incorrectly either 'x' is near 'a' (it should be 'zero') or lim f(x-a) x-->a followed by g(x)=f(x-a), which is a little backwards and doesn't follow your reasoning.

I'm sorry but I really can't tell what you're saying, not being intentionally dense, just...unintentionally dense.

 

[snipez90]

Oops, I forgot about that part about writing out the definition of the limit. So starting over, start with lim f(x) as x -> 0 = L, write out the definition, and make the obvious substitution. That is a rigorous proof, there is no need to define an extra function g.

 

[xaos]

sorry I'm a little obtuse. just follow what snipez said. start with: let f(x)-->L as x-->0 (not x-->a).

 

[zooxanthellae]

I'm still not getting this.

lim f(x) as x-->0 = l
|f(x) - l| < E
|x| < d

I'm trying to get |(x - a) - a| < d in order to get |f(x - a) - l| < E, right? But it seems like I can't get |(x - a) - a|, given that it =/= |x|. What's wrong, here?

 

[snipez90]

Well gettting 0 < |(x-a)-a| < d doesn't make much sense. In terms of the definition of the limit, writing that means you intend (x-a) to approach a, but you want x to approach a. If you don't see this, look at the definition of the limit again.

It's really quite simple, and I'm not sure if you're taking a shortcut in not writing out the full definition of the limit. At the very least you should write 0 < |x| < d IMPLIES |f(x) - L| < E. You already understand that we need to get |f(x-a) - L| < E as our conclusion. We're given 0 < |x| < d IMPLIES |f(x) - L| < E. Can you see what we need to replace x by to get from the hypothesis to the conclusion?

 

[zooxanthellae]

Well gettting 0 < |(x-a)-a| < d doesn't make much sense. In terms of the definition of the limit, writing that means you intend (x-a) to approach a, but you want x to approach a. If you don't see this, look at the definition of the limit again.

There's one of my misunderstandings. I'd been thinking about the definition as f(...) where ... as a whole approaches a, not just a single variable.

It's really quite simple, and I'm not sure if you're taking a shortcut in not writing out the full definition of the limit. At the very least you should write 0 < |x| < d IMPLIES |f(x) - L| < E. You already understand that we need to get |f(x-a) - L| < E as our conclusion. We're given 0 < |x| < d IMPLIES |f(x) - L| < E. Can you see what we need to replace x by to get from the hypothesis to the conclusion?

Well, it seems like we'd have to replace x with (x - a), but I don't see how that's possible.

 

[snipez90]

I'm not exactly sure what your misunderstanding is. I mean lim f(x-a) x --> a = L means given E > 0 there exists d > 0 such that 0 < |x-a| < d IMPLIES |f(x-a) - L | < E, which of course is just an application of the definition of the limit.

Explain to me why you think we can't replace x with x - a. One thing that's important to keep in mind is that we really do have an implication: 0 < |x| < d IMPLIES |f(x) - L| < E. This logical statement has to be taken in its entirety, and the implication cannot be ignored. I'll be happy to clarify in any case.

 

[xaos]

i think you're confusing the dummy variables in the limit.

(*) if f(y)-->L as y-->0 then f(x-a)-->L as x-->a

but what is x?

you are given several things. you are given the e,d-definition of a limit which is the hypothesis in (*) and you are given |x-a|<d in the conclusion
therefore all you need is to show is that |f(x-a)-L|<e necessarily follows.

 

[snipez90]

Good point xaos, I thought of bringing that up but I thought the implication issue was the touchy part. At least when I first encountered this type of proof, I could see that replacing x with x - a seemed to be the right approach, but I was also unsure of why I could do that. Of course you could replace x with y - a if you're not used to working with dummy variables (psychologically it's helpful at first). But at first I was still unsure because I focused too much on why I could go from 0 < |x| < d to 0 < |x-a| < d. The error of course is in not treating the implication statement in its entirety.

 

[zooxanthellae]

Let me see if I have this right: is your point that:

x as x --> 0 = (x - a) as a --> 0

and therefore substituting (x - a) for x is a valid move to make? (I think I can see how the rest follows).

I guess the main issue here might be that I don't yet have a good feel for what constitutes "rigor" vs. what constitutes "hand-waving", since the two often seem very similar to me.