Proving $\lim_{x\to1}(x^2+3)=4$ I'm learning about proving limits with the epsilon delta thing.

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

Alright, so let's grab some $\epsilon > 0$ and $\delta > 0$.
We want to prove that
$$\left|x-1 \right| < \delta \implies \left|x^2+3-4 \right| < \epsilon$$
Let's begin with
$$\left|x^2-1 \right| < \epsilon$$
$$\left|(x-1)(x+1) \right| < \epsilon$$
$$\left|(x-1)\right| \cdot\left|(x+1) \right| < \epsilon$$
Well... I have the feeling I can't simply do
$$\left|(x-1)\right| < \frac{\epsilon}{\left|(x+1) \right|}$$
What do I do in this scenario?
 A: It is okay to 'grab' the $\epsilon>0$. 
After that you go 'hunting' (not grabbing) for a suitable $\delta>0$ that makes the line written down in your question true. 
Note that $|x-1|<\delta$ leads to $|x+1|\leq3$ if $\delta$ will be taken small enough so that in these cases $|x-1||x+1|<3\delta$. 
Now choose $\delta$ in such a way that $3\delta\leq\epsilon$ and secondly that $|x+1|\leq3$ is assured. 
Here you could take $\delta=\min(\frac{1}{3}\epsilon,1)$ 
A: we want that:
$$
|x-1|<\delta \implies |x^2-1|=|x^2+3-4| <\epsilon
$$
so using a tiny trick and this fact ($|a+b|\le|a|+|b|$):
$$
|x-1|<\delta \implies |x-1|+2<\delta+2  \implies |x+1|=|x-1+2|\le |x-1|+2<\delta+2
$$
now we have:
$$
|x+1|<\delta+2 \ \ \ \ \land \ \ \ \ |x-1|<\delta
$$
so:
$$
|x^2-1|=|x-1||x+1|<\delta (\delta+2) < \epsilon
$$
now we have a criteria for choosing $\delta$ for all $\epsilon \in \mathbb{R^+}$, let's make it more explicit:
$$
\delta (\delta+2) < \epsilon \ \ \ \land \ \ \  \delta >0 \iff \boxed{0 < \delta < \sqrt{1+\epsilon}-1}
$$
so you could systematically choose $\delta$ this way:
$$
\delta = \frac{\sqrt{1+\epsilon}-1}{2}
$$
