Please help with seemingly easy limit proof

In summary: We can choose x, and then we can get u from x, but c is fixed.In summary, the conversation discusses how to prove that lim f(x) (as x goes to c) = L iff lim f(x+c) (as x goes to 0) = L. The conversation includes a suggestion to use either the sequential criterion or the definition of limit to prove this statement. The conversation also includes a detailed proof of one implication using the definition of limit. The key is to understand that x and u are arbitrary variables and represent the same concept, so the proof is valid.
  • #1
diligence
144
0

Homework Statement



Let f: R ---> R, c in R.
Show that lim f(x) (as x goes to c) = L iff lim f(x+c) (as x goes to 0) = L

Homework Equations



n/a

The Attempt at a Solution



First of all, I understand why this is true conceptually, but I'm having trouble writing a rigorous proof. I think if somebody could just help me get started on the forward direction, then i can take it from there.

There's two methods I could use: (1) I could use the sequential criterion for limits or (2) I could use the definition of the limit.

For (1), Suppose lim f(x) (as x goes to c) = L. Then by the sequential criterion, for every sequence (x(n)) contained in R such that (x(n)) converges to c, but x(n) /= c for all n, then the sequence (f(x(n))) goes converges to L. That's fine, but I'm having trouble explaining how this implies that for sequences (x(n)) that converge to 0, then (f(x(n)+c)) converges to L. Do i need to define a separate sequence converging to 0, then use the Algebraic Theorem for sequences? I don't know how to put this rigorously.

For (2), Suppose the same hypothesis as (1). Then, for all e>0, there exists a d > 0, such that if 0 < |x - c| < d, then |f(x) - L| < e. I can use the reverse triangle inequality to show that |x| < d + |c|, then define a new delta, call it d2 = d + |c|. Then i will have the condition that |x - 0| < d2. But I'm having trouble showing how this implies |f(x+c) - L| is less than some new eps, call it e2. How do I combine f(x) and c, into f(x+c)? Is there some theory of linearity that i can use for adding constants to functions defined on R?Any help with the rigorous argument would be really awesome. Which method should i use , the sequential criterion or the definition of limit? Like I said, i understand the concept completely ( i could draw a pictorial proof easily), I'm just having trouble putting it on paper into a rigorous proof that will convince my analysis instructor that I know what I'm doing :)
 
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  • #2
Try letting [itex] u = x+c [/itex] and then noting [itex] |x| = |(x+c) - c| = |u-c|[/itex].
 
  • #3
Thanks, that was really helpful! So how's this look for the forward direction?Suppose lim f(x) (as x goes to c) = L. Let e > 0. Then, by hypothesis, there exists d > 0 such that 0< |x - c| < d implies |f(x) - L| < e. Now, let u := x - c. Thus 0 < |u| < d implies |f(u + c) - L| < e. Hence, by the definition of limit, we have lim f(u +c) (as u goes to 0) = L.

Now i want to conclude that since u is arbitrary, then lim f(x+c) (as x goes to 0) = L. But can I really do that? Is u really arbitrary? It was defined very specifically, after all. I'm still having trouble understanding this subtlety. x and u are different, but they just represent arbitrary variables, so ultimately, they fundamentally represent the same thing? Is this right?
 
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  • #4
diligence said:
Thanks, that was really helpful! So how's this look for the forward direction?


Suppose lim f(x) (as x goes to c) = L. Let e > 0. Then, by hypothesis, there exists d > 0 such that 0< |x - c| < d implies |f(x) - L| < e. Now, let u := x - c. Thus 0 < |u| < d implies |f(u + c) - L| < e. Hence, by the definition of limit, we have lim f(u +c) (as u goes to 0) = L.

Now i want to conclude that since u is arbitrary, then lim f(x+c) (as x goes to 0) = L. But can I really do that? Is u really arbitrary? It was defined very specifically, after all. I'm still having trouble understanding this subtlety. x and u are different, but they just represent arbitrary variables, so ultimately, they fundamentally represent the same thing? Is this right?

why don't you plug in c = x - u? Wouldn't that work?
 
  • #5
AdrianZ said:
why don't you plug in c = x - u? Wouldn't that work?

I'm not following you, that's basically the same thing that i did, isn't it?

I'm pretty sure this proof is good (yes? no? let me know!), I'm just trying to wrap my head around the fact that saying lim f(u+c) (as u goes to 0) = L is equivalent to saying lim f(x+c) (as x goes to 0) = L, in light of the way that u was defined.
 
  • #6
diligence said:
I'm not following you, that's basically the same thing that i did, isn't it?

I'm pretty sure this proof is good (yes? no? let me know!), I'm just trying to wrap my head around the fact that saying lim f(u+c) (as u goes to 0) = L is equivalent to saying lim f(x+c) (as x goes to 0) = L, in light of the way that u was defined.

Here, I'll do one implication in detail.

Assume for all [itex] u \in \mathbb{R}[/itex] for all [itex] \epsilon >0 [/itex] there is a [itex]\delta >0[/itex] such that if [itex]0 < |u -c | < \delta [/itex] then [itex]|f(u) - L| < \epsilon[/itex].

Fix [itex] \epsilon >0 [/itex] and take [itex] \delta [/itex] as in the above assumption. Suppose [itex]x \in \mathbb{R}[/itex] with [itex] 0<|x| < \delta[/itex]. Let [itex] u = x+c[/itex]. Then [itex]|x| = |(x+c)-c| = |u-c|[/itex] so [itex]0 < |u -c | < \delta[/itex]. By assumption, then [itex] |f(u) -L| <\epsilon[/itex], so [itex] |f(x+c) -L| <\epsilon[/itex]. Since x was arbitrary, then for any x with [itex] 0<|x| < \delta[/itex], [itex] |f(x+c) -L| <\epsilon[/itex]. Therefore [itex] \lim_{x \to 0} f(x+c) = L [/itex].

The only thing that is arbitrary in the above proof is the x I picked in the first line (well, and epsilon, too).

Can you do the other direction or justify your proof now?

AdrianZ said:
why don't you plug in c = x - u? Wouldn't that work?

Maybe I'm misunderstanding you, but we don't get to choose c, it's given to us.
 
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Related to Please help with seemingly easy limit proof

1. What is a limit proof?

A limit proof is a mathematical method used to prove the behavior of a function as it approaches a specific value. It involves using algebraic manipulation and logical reasoning to show that the function approaches a certain value or does not exist at a particular point.

2. Why is this limit proof seemingly easy?

This limit proof may seem easy because it involves a simple function and a straightforward approach. However, it can still require careful reasoning and attention to detail to arrive at the correct solution.

3. What are the steps involved in this limit proof?

The steps involved in this limit proof may include identifying the function, determining the value the function is approaching, using algebraic manipulation or substitution to simplify the function, and showing that the function approaches the desired value or does not exist at the given point.

4. What are some common mistakes in limit proofs?

Some common mistakes in limit proofs include incorrect algebraic manipulation, failing to consider the behavior of the function at the given point, and not clearly stating the reasoning behind each step. It is also important to check for any restrictions on the variables in the function.

5. How can I improve my skills in limit proofs?

To improve your skills in limit proofs, it is important to practice solving various types of problems and to thoroughly understand the concepts and techniques involved. It can also be helpful to seek guidance from a teacher or tutor to identify any areas where you may be struggling.

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