It's been a while since I did a decent $\epsilon-\delta$ proof and I'm again starting my studies at university, so I am a bit rusty. I want to prove that the limit of:
$\lim_{x\rightarrow 1}x^2=1$.
Therefore I wish to show that whenever $|x-1|< \delta$, we can say that $|x^2-1|<\epsilon$. Let us find a nice $\delta$, sounds like a fun enough game! I wanted to start off by writing: $|x^2-1|= |(x-1)(x+1)|=|(x-1)||(x+1)|< \epsilon$. We thus wish that $|x-1|< \frac{\epsilon}{|x+1|} $. Now the left handside looks a whole lot like $\delta$. But we still have this nasty $x$ there. Let us restrict $\delta$, I'll say that $\delta <\frac{1}{2}$. Therefore $x \in (\frac{1}{2},\frac{3}{2})$. We can now say: $|x-1|< \frac{\epsilon}{|x+1|}<\frac{\epsilon}{\frac{1}{2}+1}<\frac{2}{3}\epsilon$. So if we let $\delta = \min (\frac{1}{2}, \frac{2}{3} \epsilon)$ we have a solid choice for $\delta$. A Proof would then be:
$\forall \epsilon>0$, let $\delta = \min (\frac{1}{2}, \frac{2}{3}\epsilon)$, then whenever $|x-1|<\delta$,
we notice $x\in (\frac{1}{2}, \frac{3}{2})$ and we also know that $|x^2-1|=|(x-1)(x+1)|=|x-1||x+1|< |x-1|\frac{3}{2}=\frac{3}{2}\delta \leq \frac{3}{2}\frac{2}{3}\epsilon=\epsilon $. Therefore $\lim_{x\rightarrow 1}x^2=1$. $\square$.
Was that good proof and are there things I can improve? While I was working on this question I figured out some stuff by myself, questions are cool like that. I feel like the last part is maybe a bit rushed, I tried to be meticulous. These proof are really quite fun if you get into them!
First, the conclusion is incorrect. If I give you $\epsilon = \frac 1{10}$ you say you can tolerate $\delta \lt \frac 2{30}$, but $|\left(\frac {28}{30}\right)^2-1|=\frac {116}{900}\approx 0.129$. When you are restricting $|x-1|$ you want to choose the most restrictive thing, which means making the right side as small as possible. Since you allow $\delta$ to otherwise be as large as $\frac 12$ you should have $|x-1| \lt \frac \epsilon{1+1.5}=\frac {2\epsilon}5$
You proof fails because you replace $x+1$ by $\frac 32$, but you want the largest value of $x+1$, which is $\frac 52$.