quick limits question

Question

Prove that $lim_{x \to 0}(x^2 -1) = -1$

$|(x^2 -1) -(-1)|< \epsilon$

$|x^2 - 1+ 1| < \epsilon$

$|x^2| < \epsilon$

$|x| < \sqrt{\epsilon}$

Can I let $\delta = \sqrt{\epsilon}$ or do I need to restrict $\delta$ here?

Answer 1

Indeed, $\delta = \sqrt \varepsilon$ is a suitable choice. We can check this as well.

Let $|x-0| < \delta$, i.e. $|x| < \sqrt \varepsilon$. We wish to show, on this premise, that $|(x^2-1) - (-1)| < \varepsilon$; equivalently, that $|x^2| < \varepsilon$. We have no issue in showing this, as, per a property of the absolute value,

$$|x^2| = |x|^2 < (\sqrt \varepsilon)^2 = \varepsilon$$

which verifies your choice of $\delta$.