Epsilon-Delta More Confusion

Question

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?

Answer 1

$$(x^2 + 3) - 4 = (x - 1)(x + 1) $$

First choose $\delta= 1$ so that

$x + 1 = (x - 1) + 2 \lt x - 1 + 2 \lt 1 + 2 = 3$.

Then let $\epsilon> 0$ be given, choose $\delta = \min\{\epsilon/3, 1\}$, then if $x - 1 \lt \delta \implies (x^2 + 3) - 4 \lt \epsilon$.

Answer 2

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 $\delta$?