Proof of Limits

Hi, I have the following question:



I did the proof as dictated by a book, and have the following:





Could someone please check if it's ok, because there is no solutions in the book. Also I'll be uploading other exercises after this so just please check if it's ok

The mistake I noticed was that the final line in the 1st image should use the fact that

$\displaystyle -\epsilon - 4 < -x < \epsilon - 4 \Rightarrow \epsilon + 4 > x > 4 -\epsilon$

Also, it's always better to use quantifiers, at undergraduate level anyway.

quantifiers?

$\forall$

means "for all"

$\exists$

means "there exists"

As the name suggests, they relate to quantity.

Could someone please check if it's ok, because there is no solutions in the book. Also I'll be uploading other exercises after this so just please check if it's ok

It is kind of ok (a couple of slip ups) but here is the main problem. The order and the way you have written it out is horrendous.

When you are writing an epsilon delta type proof, you should start off by fixing epsilon and defining delta and showing that it works.

What the book has told you is the way in general to find a delta that works. What you have written out is your workings out/scrap notes that you would calculate BEFORE writing out an actual answer. When you write out an answer, you do it in the way easiest to read and follow, not in the order following the thought process needed to get to it.

In this case, you have overcomplicated something extremely simple. For example, here is my proof.

Let $\epsilon > 0$ be given and define $\delta := \epsilon$.

Then $0 < |x-4| < \delta$ is equivalent to $0 < |4-x| < \delta$ (by definition of the modulus) which is further equivalent to $0 < |4-x| < \epsilon$ (since $\epsilon = \delta$)

The final inequality may be written as $0 < |f(x)-L| <\epsilon$

where $f(x) = 9-x$ and $L=5$.

Therefore, by definition, the limit of $f(x)$ as $x \rightarrow 4$ is $5$.


The point was that both inequalities were the same with epsilon replaced with delta so choosing delta equal to epsilon makes the RHS equivalent to the LHS (which is even stronger than what you need).

but then where do I go from here?

I don't think you understood what I was saying. The game with these proofs is that if I give you an $\epsilon >0$, you give me a $\delta$ such that if $x$ is within $\pm \delta$ of a point $c$, then $f(x)$ is within $\pm \epsilon$ of $L$.

How do I know what the value of x will be?

You are considering a limit at a particular point inside the domain at your function. The values of x are the ones in some neighbourhood of that point. Go back and read the definition of a limit and try to understand what it is...

You are showing that you can get the value of a function as close as you like to the limit (lying at a distance of $\pm \epsilon$ by taking any value of $x$ within $\pm \delta$ of the point $c$. Try to think of the picture.

You really have to take the time to read, learn and inwardly digest the definition of a limit before you start being able to work out and write a proof that a limit exists. Otherwise, you will get very confused by what the book (or me or whoever) is actually trying to do.

You are considering a limit at a particular point inside the domain at your function. The values of x are the ones in some neighbourhood of that point. Go back and read the definition of a limit and try to understand what it is...

You are showing that you can get the value of a function as close as you like to the limit (lying at a distance of $\pm \epsilon$ by taking any value of $x$ within $\pm \delta$ of the point $c$. Try to think of the picture.

You really have to take the time to read, learn and inwardly digest the definition of a limit before you start being able to work out and write a proof that a limit exists. Otherwise, you will get very confused by what the book (or me or whoever) is actually trying to do.

I end up with two values.. I tried setting one, but I don't end up with
|f(x)-L|<epsilon , how do I set a value?

To be honest, I can't really understand what you wrote... for instance I don't know why you would make delta be a function of x...

But if you go back to my previous post, you will see that I found that if I assume that delta is less than one, I could then show that if delta was a particular multiple of epsilon, it would work. In that situation therefore, I would define delta to be the minimum of 1 and the particular multiple of epsilon that I determined so that both conditions would be satisfied. Without looking at what you have done, try taking delta to be the minimum of the two values and see if you can write out a proper proof (in here if you like) and we can help check it or otherwise, you will get to a point where it won't work and you can hopefully see the problem.

ook... how about this -->