Help with Epsilon-Delta proof for limits

Question

I tried to teach myself these types of proofs. I understand the reasoning behind it very well, but I have trouble understanding specific parts when simplifying inequalities. Let me give an example:

Say I wanted to prove the following:

$$\lim_{x\to1}(x^2+3)=4$$

I start by supposing: given $ε>0$, I want to find $δ>0$ such that

$$0<|x-1|<δ => |(x^2+3)-4|<ε$$

I start by simplifying the RHS to find an expression that relates $ε$ to $δ$:

$$|x^2-1|<ε$$ $$=>|(x-1)(x+1)|<ε$$ $$=>|x-1||x+1|<ε$$

At this point I was stuck and did research on how I should proceed. Apparently we can restrict $δ$ to only be at most $1$ unit away from $a$. Since we are dealing with the limit of $f(x)$ as ${x\to a}$, it is reasonable to restrict our "radius" around $a$ this way. I sort of understand this reasoning, although a more detailed explanation would be appreciated. Anyway, this implies (in my case):

$$|x-1|<δ≤1$$ $$=>|x-1|<1$$ $$=>0<x<2$$ $$=>1<x+1<3$$ $$=>1<|x+1|<3$$

This means that the min value of |x+1| is larger than 1 and the max value is smaller than 3. However, here comes the part where I get stuck: according to multiple solutions I found online, it is reasonable to say that:

$$|x-1||x+1|<3|x-1|$$

Now the part above I totally understand, but the following part I do not. Apparently, it is a logical step to deduce the following:

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$$|x-1||x+1|<3|x-1|<ε$$ $$=>3|x-1|<ε$$

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How can we logically deduce that $$3|x-1|$$ is smaller than the given $ε$? In my reasoning, if $$3|x-1|$$ is larger than $$|x+1||x-1|$$ it does not necessarily mean that the former is also smaller than $ε$. For instance, if $3<5$ and $6>3$, then it does NOT mean that $6<5$, obviously. In my opinion, it is correct to deduce the following:

$$|x-1|<|x+1||x+1|$$

(Using the same reasoning as earlier)

$$=>|x-1|<|x+1||x-1|<ε$$ $$=>|x-1|<ε$$

This, to me, is pretty clear. Any expression smaller than the middle one is logically smaller than the third. Therefore, I set ε=δ. But then I'm stuck again. In the proofs I looked at online, they say to set $$δ=min{(1,ε)}$$, to pick the smaller value of the two. Why is that?

To sum up, I would appreciate any feedback on my work, especially an explanation on why the part between ****(...)**** is a valid deduction and why we pick the smallest value for $δ$? Thanks!

Answer 1

"How can we logically conclude that $3|x-1|<\epsilon$?"

We don't. We conclude that as long as $\delta<1$ , then $|x+1|\cdot |x-1|<3|x-1|$. We want $|x+1|\cdot |x-1|<\epsilon$, so if we can somehow make sure that $3|x-1|<\epsilon$, then we are good.

So now we are free to set our $\delta$ to make sure that the last inequality above holds. Choosing any $\delta<\epsilon/3$ should work nicely, since if $|x-1|<\delta$, then $$ |x^2-1|=|x+1|\cdot|x-1|< 3|x-1|<3\delta < \epsilon $$ But remember that the middle step here also required $\delta<1$, so actually we have to pick $\delta<\min(1,\epsilon/3)$.

Answer 2

We get to choose our $\delta$, hence to get the conclusion that

$$3|x-1|<\epsilon$$

I can choose my $\delta$ to be $$\delta=\min(1, \epsilon/3)$$

Hence, For $|x-1| <\delta$,

$$|(x+1)(x-1)|<3|x-1|<3\delta\leq 3(\epsilon/3)=\epsilon$$

$|x+1|<3$ is due to $\delta \leq 1$, $3\delta \leq \epsilon$ is due to we choose $\delta \leq \epsilon/3$.

Since I want $\delta$ to be both smaller than $1$ and $\epsilon/3$, I choose the minimal of both quantities.

Answer 3

If you start from $\epsilon$ (as you have done and as it's correct in general, if we have not an intuition about $\delta$), than you have to solve the inequality: $$ |x^2-1|<\epsilon $$

this is equivalent to the two systems: $$ \begin{cases} x^2-1\ge0\\ x^2<1+\epsilon \end{cases} \quad \lor \quad \begin{cases} x^2-1<0\\ x^2>1-\epsilon \end{cases} $$ With the condition $x>0$ (since we have $x \to 1$), the solution of this is:

$$ [1,\sqrt{1+\epsilon}) \cup(\sqrt{1-\epsilon},1)=(\sqrt{1-\epsilon},\sqrt{1+\epsilon}) $$

so you can chose $\delta= \min\{1-\sqrt{1-\epsilon},\sqrt{1+\epsilon}-1\}$