How does $ax^2 − 2x^2 − a^2x − 2a^2 $ result in $ |ax||x − a| + 2|x^2 − a^2 | $ by the triangle inequality?

Question

This is for part of an epsilon-delta proof showing that the limit as $x\to a$ of $ \frac{x^2}{(x-2)}=\frac{a^2}{(a-2)}$...the solution for this section gave me this. I am assuming it is using the triangle inequality, though applying it with both variables for some reason messes with me.

Answer 1

Yes, it is by the triangle inequality. But you also made a mistake in the expression in the title of your question. After subtracting those two fractions, you will have $$ax^2−2x^2−\color{red}{(}a^2x−2a^2\color{red}{)}=ax^2−2x^2−a^2x\color{red}{+}2a^2.$$ The lesson here is: use parentheses for grouping, because you may need to distribute something!

Then, for this epsilon-delta proof one needs to estimate $|ax^2−2x^2−a^2x+2a^2|$. Note that $$|ax^2−2x^2−a^2x+2a^2|=|(ax^2−a^2x)−(2x^2-2a^2)|\le|ax^2−a^2x|+|2x^2-2a^2|$$ by the triangle inequality.