The question I am trying to answer is "Prove that for all $\varepsilon \in (0, 1)$, if $|x − 2| < \frac{\varepsilon}{7}$ then $|x^2 − 4| < \varepsilon$."
My attempt is the following:
Suppose that $\varepsilon \in (0, 1)$. Let $|x-2|< \frac{\varepsilon}{7}$. Now consider \begin{align*} |x^2 - 4|&=|x-2||x+2|\\ &=|x-2||x-2+4|\\ &< \frac{\varepsilon}{7}\left(\frac{\varepsilon}{7} + 4\right) \\ &<7 \frac{\varepsilon}{7} \\ &= \varepsilon. \end{align*} QED
However I'm not too sure whether the last two lines are correct. Any help would be very appreciated. Thank you.
They are valid. You just used the triangular inequality as $$|x+2|=|x-2+4|\leq |x-2|+4<\frac{\varepsilon}{7}+4<\frac{1}{7}+4<7$$ so, indeed, $|x^2-4|=|x-2|\cdot|x+2|<\frac{\varepsilon}{7}\cdot7=\varepsilon$.