Where in this statement is the reverse triangle rule used?

Question

I've attached part of a proof from my lecture notes (the proof is showing that $\frac{1}{x^2}$ is continuous over $\mathbb{R} \setminus \{0\}$), it makes reference to using the reverse triangle rule in an important step, but I really don't see what it is being used on.

Answer 1

By hypothesis in red

$$\lvert x - a \rvert \lt \delta \le \frac{\lvert a \rvert}{2}.$$

Hence using the reverse triangle inequality

$$\lvert \lvert x \rvert - \lvert a \rvert \rvert \le \frac{\lvert a \rvert}{2}$$ which implies

$$-\frac{\lvert a \rvert}{2} \le \lvert x \rvert - \lvert a \rvert \le \frac{\lvert a \rvert}{2}$$ and finally the desired conclusion.

Answer 2

If $|x-a|<\frac{|a|}2$, then $\bigl||x|-|a|\bigr|<\frac{|a|}2$. So:

• since $|x|-|a|\leqslant\bigl||x|-|a|\bigr|$, $|x|-|a|<\frac{|a|}2$;
• since $|a|-|x|<\bigl||x|-|a|\bigr|$, $|a|-|x|<\frac{|a|}2$.

Therefore$$\frac{|a|}2<|x|<\frac{3|a|}2.$$