Let $f(x) = x^2$ and let $\varepsilon > 0$ be given. Find a $\delta$ so that $|x-1|<\delta$ implies $|f(x) - 1|<\epsilon$.
I have $|x^2 - 1| < \varepsilon \implies |(x+1)(x-1)| < \varepsilon$, which becomes $|x-1| \times |x+1| < \varepsilon$, but I am not sure where this goes next.
