Stuck on an epsilon-delta proof where I let delta be a minimum of two values I'm currently studying for my real analysis 1 midterm Thursday and I'm trying to learn epsilon-delta proofs. Unfortunately, my instructor has decided to use a custom textbook that has no examples in it and he uses class time for group discussions instead of actually lecturing. I'm trying to solve delta-epsilon proofs I find online, including this website, but I have a question about writing proofs that involve letting delta be the minimum of two options.
Here is the problem I found online that I'm currently stuck on:
Prove that x$^2$ + 3 goes to 4 as x goes to 1.
Here's my proof:
Assume $\epsilon$ > 0 is given. Let $\delta$ = min(1, $\frac{\epsilon}{3}$). Since $\epsilon$ > 0, we also have that $\delta$ > 0. Now, for every x, 0 < |x - a| < $\delta$ implies |x - 1| < ...
Now this is where I'm unsure of how to proceed. The website I found the problem on has hints to construct the delta so that's how I was able to find that, but then it does not tell you how to move forward to prove the limit. Am I supposed to include two cases here, one when $\delta$ = 1 and another when $\delta$ = $\epsilon$/3, or can I just work forward with the delta involving epsilon? If anyone would be so kind as to clear this bit of confusion up for me, I'm confident I can prove the limit with no troubles. I'm just unsure of how to proceed at this specific step because I can't seem to find a worked example of such a case that covers more than finding the delta in such a situation.
If anyone would be willing to clear this up for me (one line saying "do two cases" or "use the epsilon" would be all I need), I'd be greatly appreciative. Thank you.
 A: Let $\epsilon >0$. Choose $\delta =\min{(1,\epsilon/3)}$. Suppose that
$0 < |x-1| < \delta$. What we need to show is that under these conditions,
$| (x^2+3)-4 | < \epsilon $. In other words if we show $ |x^2-1| < \epsilon $ we are done. Well, we have that,
\begin{align}
|x^2-1| & =|(x-1)(x+1)| \\
& < \delta |(x-1)+2|, \quad \text{since}\; |x-1| < \delta \; \text{by assumption} \tag{A}. \\
&  \leq \delta  \cdot (|(x-1)|+2), \quad \text{using triangle inequality} \\
& \leq \delta \cdot (1+2), \quad \text{since}\; |x-1| < \delta \leq 1\\
& =3\delta \\
& \leq 3 \left( \frac{\epsilon}{3} \right), \quad \text{since}\; \delta \leq \epsilon/3 \\
& =\epsilon.
\end{align}
And so we are done. The trick in (A) was to write $ (x+1)$ as $(x-1+2)$ and then using the triangle inequality. I hope this clears it up.
EDIT: Choosing $\delta = \min{(1,\epsilon/3)}$ implies that $ \delta \leq 1$ AND $\delta \leq \epsilon/3$. Notice that we are NOT choosing $\delta =1$ or $\delta =\epsilon/3$ separately. We are choosing $\delta$ to be either $1$ or $\epsilon/3$ whichever is smaller ( We don't know which one is smaller a priori). However this choice of $\delta$ implies the above two inequalities: $\delta \leq 1$ AND $\delta  \leq \epsilon/3$. These two inequalities being true at the same time are essential to our proof.
Hope this helps. 
A: I think it would do benefit you to look at this alternative approach instead, and see how it works for you. What you need to prove is that $$\forall{\epsilon\gt0},\exists{\delta\gt0},\forall{x\in\mathbb{R}},\left[0\lt|x-1|\lt\delta\implies|(x^2+3)-4|\lt\epsilon\right].$$ Notice that $|(x^2+3)-4|=|x^2-1|=|(x+1)(x-1)|=|x+1||x-1|=|(x-1)+2||x-1|$. By the triangle inequality, $|(x-1)+2|\leq|x-1|+2$, hence $|(x-1)+2||x-1|\leq|x-1|^2+2|x-1|$. Thus $$\forall{x\in\mathbb{R}},\left[|x-1|\lt\delta\implies|(x^2+3)-4|\leq|x-1|^2+2|x-1|\lt\delta^2+2\delta\right].$$ We want $|(x^2+3)-4|\lt\delta^2+2\delta$ to be equivalent to $|(x^2+3)-4|\lt\epsilon$. This can be accomplished by letting $\epsilon=\delta^2+2\delta$. The idea is then to prove that $$\forall{\epsilon\gt0},\exists{\delta\gt0},\epsilon=\delta^2+2\delta.$$ This can indeed be proven, because $\epsilon=\delta^2+2\delta$ is equivalent to $\epsilon+1=\delta^2+2\delta+1=(\delta+1)^2$, hence $\sqrt{\epsilon+1}=|\delta+1|$, and the requirement that $\delta\gt0$ forces $|\delta+1|=\delta+1$, so $\delta=\sqrt{\epsilon+1}-1$, and indeed, $$\forall{\epsilon\gt0},\delta=\sqrt{\epsilon+1}-1\gt0.$$ Therefore, $$\forall{\epsilon\gt0},\exists{\delta=\sqrt{\epsilon+1}-1\gt0},\forall{x\in\mathbb{R}},\left[0\lt|x-1|\lt\delta=\sqrt{\epsilon+1}-1\implies0\lt|x+1||x-1|=|(x^2+3)-4|\lt|x+1|\delta\lt(\delta+2)\delta=\left(\sqrt{\epsilon+1}+1\right)\left(\sqrt{\epsilon+1}-1\right)=\epsilon\right].$$ That finishes the proof. I think this approach is more intuitive and natural, as one need not get into contrived assumptions about $\delta$ and one need not utilize the minimum or maximum functions, as those are rather complicated to use for these proofs.
