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 can factor $x^2-1$. From there, the condition $|x-1|<\delta$ implies that $|x+1|<\delta +2$. Can you use this to find an expression for $\delta$?
