Epsilon-Delta More Confusion 
Use Epsilon-Delta to prove:
  $$
\lim_{x \to 1} (x^2 + 3) = 4
$$

So, we need to find a $\delta$ s.t. 
$$
0 < x - 1 < \delta \; \implies \; 0 < |(x^2 + 3) - 4| < \epsilon
$$
We simplify
$$0< |(x^2 + 3) - 4| < \epsilon$$ to get $$0 < |x^2 - 1| < \epsilon$$
This is where I'm stuck. 
How do I find a delta in terms of epsilon now?

You want to find a function $f$ so that for any $e > 0$
$$
|x - 1| < f(e) \implies |(x^2 + 3) - 4| = |x^2 - 1| = |x + 1||x - 1| < e
$$
Since $|x - 1| < f(e) \implies |x + 1| < f(e) + 2$ you also have a
bound for the product of those factors
$$
|x + 1||x - 1| < f(e)(f(e) + 2)
$$
Now all you have to do is set $f(e)(f(e) + 2) = e$ and solve for
$f(e)$:
$$
f(e)^2 + 2f(e) = e \iff (f(e) + 1)^2 = e + 1 \iff f(e) = \sqrt{e + 1} - 1
$$
So $\delta = f(e) = \sqrt{e + 1} - 1$. As you can see, there are many ways to choose the $\delta$. I like this way because the choice feels less arbitrary.
