# Thread: Yet another epsilon-delta problem?

1. ## Yet another epsilon-delta problem?

Okay, I think I have this problem though. I just need somebody to verify weather my methodology is correct. Heres the problem:

Prove that:

$\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$

Heres my thought, since the left side limit, by its definition, the $\displaystyle x$ in the $\displaystyle f(x)$ approaches 0. So, I figured (and this is where I needed some help verifiying this assumption) I figured that all I needed to do is show that $\displaystyle (x-a)$ approaches 0 since its the same function $\displaystyle f(x)$. Does that reasoning make sense? And if that reasoning it valid, then we have:

$\displaystyle |x-a-0| = |x-a| < \delta$

thus if we set $\displaystyle \delta = \epsilon$ then it shows that $\displaystyle x-a$ approaches $\displaystyle 0$ as $\displaystyle x$ approaches $\displaystyle a$, right?

2. Is there any additional information in the question? Does the question mention that the left limit exists? Or are you asked to show that both sides behave the same?

Assuming that the left limit exists, I wrote out the definition of the limit and got this. Given epsilon > 0, there is a d such that |x-0| < d => |f(x)-L|< e, where L is the assumed limit. Here I just make a change of variables x = y - a and then plug it in. That's equivalent to the right hand definition. I think it should be straight-forward to show that if the left side doesn't exist, then so does the right hand side.

3. I really fail to see how this thread differs from this.

Aren't the ideas the same?

4. Oh, well, there you go!

5. Originally Posted by Plato
I really fail to see how this thread differs from this.

Aren't the ideas the same?
It differs large enough to somebody who is new to the formal definition of a limit. Obviously, the best way too actually increase my understanding of epsilon-delta proofs is to make sure I'm actually doing the proofs correctly. When my textbook lacks the answer to a specific problem, and I would like to find out the correct answer, I consult this forum. I got this textbook early, to study over the summer before classes this fall. As a consequence I don't have any teachers or fellow students to consult, mainly in terms of checking the validity of my proofs. All I have is this forum. So, I apologize if this thread question seems exceedingly redundant, but consider this: even the slightest of differences in a specific problem, at this premature, introductory level of epsilon-delta proofs, the level that I'm at now; even the slightest difference can bring cause for personal concern, or a disconnect in the neccesarry understanding of the proof involved. I'm sure that to most people on this forum, the two posts I've recently made about epsilon-delta proofs will probably seem incredibly alike. But just understand that even with the simmilarities involved, I still wasn't certain that my proof was valid for this simmilar problem. I'm still just learning the basics of epsilon-delta proofs. I'm not trying to be some annoying, double posting idiot. I'm just trying to do my best to get ahead of learning this stuff before fall, and this forum is my only resource as of now.

6. Originally Posted by mfetch22
It differs large enough to somebody who is new to the formal definition of a limit. Obviously, the best way too actually increase my understanding of epsilon-delta proofs is to make sure I'm actually doing the proofs correctly. When my textbook lacks the answer to a specific problem, and I would like to find out the correct answer, I consult this forum. I got this textbook early, to study over the summer before classes this fall. As a consequence I don't have any teachers or fellow students to consult, mainly in terms of checking the validity of my proofs. All I have is this forum. So, I apologize if this thread question seems exceedingly redundant, but consider this: even the slightest of differences in a specific problem, at this premature, introductory level of epsilon-delta proofs, the level that I'm at now; even the slightest difference can bring cause for personal concern, or a disconnect in the neccesarry understanding of the proof involved. I'm sure that to most people on this forum, the two posts I've recently made about epsilon-delta proofs will probably seem incredibly alike. But just understand that even with the simmilarities involved, I still wasn't certain that my proof was valid for this simmilar problem. I'm still just learning the basics of epsilon-delta proofs. I'm not trying to be some annoying, double posting idiot. I'm just trying to do my best to get ahead of learning this stuff before fall, and this forum is my only resource as of now.
Epsilon-delta proofs can be really tricky, even if you get them 100%. I feel for you, because this probably the most difficult part of the introductory calculus because it is just so abstract. It'll help if you really write things out at first and be exact to the point when using the definitions. Btw, does my post make sense?

7. Originally Posted by Vlasev
Epsilon-delta proofs can be really tricky, even if you get them 100%. I feel for you, because this probably the most difficult part of the introductory calculus because it is just so abstract. It'll help if you really write things out at first and be exact to the point when using the definitions. Btw, does my post make sense?
Yes, your post makes sense. But my question was less about "How do you prove this limit?" and was more about "Is this line of reasoning correct?". Please, consider this line of reasoning and let me know if it is valid, because then I will know if I can use it on future problems. This is what we started with (and yea, this is exactly, and only, what the textbook gave, so I guess we assume the limit exists):

Prove:

$\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$

If you look at the left side of the equation, the notation exactly states that "x approaches a", so that is something we can take as given right? Now heres the line of reasoning I was trying to validate:

Consider the values $\displaystyle f(a)$ and $\displaystyle f(b)$. Obviously, if we have:

$\displaystyle a=b$

then

$\displaystyle f(a) = f(b)$

since $\displaystyle f$ denotes the same function. So, on top of that, consider the following limits:

$\displaystyle \lim_{x \to a} f(b)$ and $\displaystyle \lim_{x \to a} f(c)$

now, it follows that if we have:

$\displaystyle b = c$

then

$\displaystyle f(b) = f(c)$

then

$\displaystyle \lim_{x \to a} f(b) = \lim_{x \to a} f(c)$

since the same function $\displaystyle f(b)=f(c)$ produces the same limits, by the theorem that shows any given functions approaches only one limit a a specified point, then the limits are equal. Granted, my demonstration above is not "rigourous", and it slightly "abuses" notation, but I think you get my point. Now, given the above, we return to the problem of proving:

$\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$

By the investigations above, I beleive it follows that we need only to prove that:

$\displaystyle f(x) \to f(0)$ and $\displaystyle f(x-a) \to f(0)$

when

$\displaystyle x \to 0$ and $\displaystyle x \to a$

, respectively. But, to show that, it follows that we need only to prove that

$\displaystyle (x - a) \to 0$,

since by the definition of a limit, the $\displaystyle x$ on the left side approaches 0, we simply need to show the argument of both

$\displaystyle f(x)$ and $\displaystyle f(x-a)$

to approach the same value. Is this reasoning valid?

And from that, and the definition of a limit, and the information on the left side, we start with:

$\displaystyle 0 < |x-0| = |x| < \delta_1$

So we know the argument of $\displaystyle f(x)$ on the left side approaches 0. And as for the right side, we need to show that if $\displaystyle x \to a$ then $\displaystyle (x-a) \to 0$. We start this by:

$\displaystyle 0 < |x-a| < \delta_2$

to show $\displaystyle (x-a) \to 0$ we need to find some $\displaystyle \epsilon > 0$ such that:

$\displaystyle |(x-a) - 0| < \epsilon$

but obviously we have:

$\displaystyle |(x-a)-0| = |x-a| < \delta_2$

thus, if set $\displaystyle \delta_2 = \epsilon$ we have shown that, for all $\displaystyle \delta_2 > 0$ and $\displaystyle \epsilon > 0$ that:

$\displaystyle 0 < |x-a| < \delta_2$

implies

$\displaystyle |(x-a)-0| < \epsilon$

by the definition of a limit, we have:

$\displaystyle \lim_{x \to a} (x-a) = 0$

and it follows that:

$\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$

Phew! That was allot, now, my question, is this a valid "line-of-reasoning", and consequently, a valid proof?